Related
I came across this question from a colleague.
Q: Given a huge list (say some thousands)of positive integers & has many values repeating in the list, how to find those values occurring odd number of times?
Like 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 1 2 3 4 5 6 1 2 3 4 5 1 2 3 4 1 2 3 1 2 1...
Here,
1 occrus 8 times
2 occurs 7 times (must be listed in output)
3 occurs 6 times
4 occurs 5 times (must be listed in output)
& so on... (the above set of values is only for explaining the problem but really there would be any positive numbers in the list in any order).
Originally we were looking at deriving a logic (to be based on c).
I suggested the following,
Using a hash table and the values from the list as an index/key to the table, keep updating the count in the corresponding index every time when the value is encountered while walking through the list; however, how to decide on the size of the hash table?? I couldn't say it surely though it might require Hashtable as big as the list.
Once the list is walked through & the hash table is populated (with the 'count' number of occurrences for each values/indices), only way to find/list the odd number of times occurring value is to walk through the table & find it out? Is that's the only way to do?
This might not be the best solution given this scenario.
Can you please suggest on any other efficient way of doing it so??
I sought in SO, but there were queries/replies on finding a single value occurring odd number of times but none like the one I have mentioned.
The relevance for this question is not known but seems to be asked in his interview...
Please suggest.
Thank You,
If the values to be counted are bounded by even a moderately reasonable limit then you can just create an array of counters, and use the values to be counted as the array indices. You don't need a tight bound, and "reasonable" is somewhat a matter of platform. I would not hesitate to take this approach for a bound (and therefore array size) sufficient for all uint16_t values, and that's not a hard limit:
#define UPPER_BOUND 65536
uint64_t count[UPPER_BOUND];
void count_values(size_t num_values, uint16_t values[num_values]) {
size_t i;
memset(count, 0, sizeof(count));
for (i = 0; i < num_values; i += 1) {
count[values[i]] += 1;
)
}
Since you only need to track even vs. odd counts, though, you really only need one bit per distinct value in the input. Squeezing it that far is a bit extreme, but this isn't so bad:
#define UPPER_BOUND 65536
uint8_t odd[UPPER_BOUND];
void count_values(size_t num_values, uint16_t values[num_values]) {
size_t i;
memset(odd, 0, sizeof(odd));
for (i = 0; i < num_values; i += 1) {
odd[values[i]] ^= 1;
)
}
At the end, odd[i] contains 1 if the value i appeared an odd number of times, and it contains 0 if i appeared an even number of times.
On the other hand, if the values to be counted are so widely distributed that an array would require too much memory, then the hash table approach seems reasonable. In that case, however, you are asking the wrong question. Rather than
how to decide on the size of the hash table?
you should be asking something along the lines of "what hash table implementation doesn't require me to manage the table size manually?" There are several. Personally, I have used UTHash successfully, though as of recently it is no longer maintained.
You could also use a linked list maintained in order, or a search tree. No doubt there are other viable choices.
You also asked
Once the list is walked through & the hash table is populated (with the 'count' number of occurrences for each values/indices), only way to find/list the odd number of times occurring value is to walk through the table & find it out? Is that's the only way to do?
If you perform the analysis via the general approach we have discussed so far then yes, the only way to read out the result is to iterate through the counts. I can imagine alternative, more complicated, approaches wherein you switch numbers between lists of those having even counts and those having odd counts, but I'm having trouble seeing how whatever efficiency you might gain in readout could fail to be swamped by the efficiency loss at the counting stage.
In your specific case, you can walk the list and toggle the value's existence in a set. The resulting set will contain all of the values that appeared an odd number of times. However, this only works for that specific predicate, and the more generic count-then-filter algorithm you describe will be required if you wanted, say, all of the entries that appear an even number of times.
Both algorithms should be O(N) time and worst-case O(N) space, and the constants will probably be lower for the set-based algorithm, but you'll need to benchmark it against your data. In practice, I'd run with the more generic algorithm unless there was a clear performance problem.
Random question.
I am attempting to create a program which would generate a pseudo-random distribution. I am trying to find the right pseudo-random algorithm for my needs. These are my concerns:
1) I need one input to generate the same output every time it is used.
2) It needs to be random enough that a person who looks at the output from input 1 sees no connection between that and the output from input 2 (etc.), but there is no need for it to be cryptographically secure or truly random.
3)Its output should be a number between 0 and (29^3200)-1, with every possible integer in that range a possible and equally (or close to it) likely output.
4) I would like to be able to guarantee that every possible permutation of sequences of 410 outputs is also a potential output of consecutive inputs. In other words, all the possible groupings of 410 integers between 0 and (29^3200)-1 should be potential outputs of sequential inputs.
5) I would like the function to be invertible, so that I could take an integer, or a series of integers, and say which input or series of inputs would produce that result.
The method I have developed so far is to run the input through a simple halson sequence:
boost::multiprecision::mpz_int denominator = 1;
boost::multiprecision::mpz_int numerator = 0;
while (input>0) {
denominator *=3;
numerator = numerator * 3 + (input%3);
input = input/3;
}
and multiply the result by 29^3200. It meets requirements 1-3, but not 4. And it is invertible only for single integers, not for series (since not all sequences can be produced by it). I am working in C++, using boost multiprecision.
Any advice someone can give me concerning a way to generate a random distribution meeting these requirements, or just a class of algorithms worth researching towards this end, would be greatly appreciated. Thank you in advance for considering my question.
----UPDATE----
Since multiple commenters have focused on the size of the numbers in question, I just wanted to make clear that I recognize the practical problems that working with such sets poses but in asking this question I'm interested only in the theoretical or conceptual approach to the problem - for example, imagine working with a much smaller set of integers like 0 to 99, and the permutations of sets of 10 of output sequences. How would you design an algorithm to meet these five conditions - 1)input is deterministic, 2)appears random (at least to the human eye), 3)every integer in the range is a possible output, 4)not only all values, but also all permutations of value sequences are possible outputs, 5)function is invertible.
---second update---
with many thanks to #Severin Pappadeux I was able to invert an lcg. I thought I'd add a little bit about what I did to hopefully make it easier for anyone seeing this in the future. First of all, these are excellent sources on inverting modular functions:
https://www.khanacademy.org/computing/computer-science/cryptography/modarithmetic/a/modular-inverses
https://www.khanacademy.org/computer-programming/discrete-reciprocal-mod-m/6253215254052864
If you take the equation next=ax+c%m, using the following code with your values for a and m will print out the euclidean equations you need to find ainverse, as well as the value of ainverse:
int qarray[12];
qarray[0]=0;
qarray[1]=1;
int i =2;
int reset = m;
while (m % a >0) {
int remainder=m%a;
int quotient=m/a;
std::cout << m << " = " << quotient << "*" << a << " + " << remainder << "\n";
qarray[i] =qarray[i-2]-(qarray[i-1]*quotient);
m=a;
a=remainder;
i++;
}
if (qarray[i-1]<0) {qarray[i-1]+=reset;}
std::cout << qarray[i-1] << "\n";
The other thing it took me a while to figure out is that if you get a negative result, you should add m to it. You should add a similar term to your new equation:
prev = (ainverse(next-c))%m;
if (prev<0) {prev+=m;}
I hope that helps anyone who ventures down this road in the future.
Ok, I'm not sure if there is a general answer, so I would concentrate on random number generator having, say, 64bit internal state/seed, producing 64bit output and having 2^64-1 period. In particular, I would look at linear congruential generator (aka LCG) in the form of
next = (a * prev + c) mod m
where a and m are primes to each other
So:
1) Check
2) Check
3) Check (well, for 64bit space of course)
4) Check (again, except 0 I believe, but each and every permutation of 64bits is output of LCG starting with some seed)
5) Check. LCG is known to be reversible, i.e. one could get
prev = (next - c) * a_inv mod m
where a_inv could be computed from a, m using Euclid's algorithm
Well, if it looks ok to you, you could try to implement LCG in your 15546bits space
UPDATE
And quick search shows reversible LCG discussion/code here
Reversible pseudo-random sequence generator
In your update, "appears random (to the human eye)" is the phrasing you use. The definition of "appears random" is not a well agreed upon topic. There are varying degrees of tests for "randomness."
However, if you're just looking to make it appear random to the human eye, you can just use ring multiplication.
Start with the idea of generating N! values between 0 and M (N>=410, M>=29^3200)
Group this together into one big number. we're going to generate a single number ranging from 0 to *M^N!. If we can show that the pseudorandom number generator generates every value from 0 to M^N!, we guarantee your permutation rule.
Now we need to make it "appear random." To the human eye, Linear Congruent Generators are enough. Pick a LCG with a period greater than or equal to 410!*M^N satisfying the rules to ensure a complete period. Easiest way to ensure fairness is to pick a LCG in the form x' = (ax+c) mod M^N!
That'll do the trick. Now, the hard part is proving that what you did was worth your time. Consider that the period of just a 29^3200 long sequence is outside the realm of physical reality. You'll never actually use it all. Ever. Consider that a superconductor made of Josephine junctions (10^-12kg processing 10^11bits/s) weighing the mass of the entire universe 3*10^52kg) can process roughly 10^75bits/s. A number that can count to 29^3200 is roughly 15545 bits long, so that supercomputer can process roughly 6.5x10^71 numbers/s. This means it will take roughly 10^4600s to merely count that high, or somewhere around 10^4592 years. Somewhere around 10^12 years from now, the stars are expected to wink out, permanently, so it could be a while.
There are M**N sequences of N numbers between 0 and M-1.
You can imagine writing all of them one after the other in a (pseudorandom) sequence and placing your read pointer randomly in the resulting loop of N*(M**N) numbers between 0 and M-1...
def output(input):
total_length = N*(M**N)
index = input % total_length
permutation_index = shuffle(index / N, M**N)
element = input % N
return (permutation_index / (N**element)) % M
Of course for every permutation of N elements between 0 and M-1 there is a sequence of N consecutive inputs that produces it (just un-shuffle the permutation index). I'd also say (just using symmetry reasoning) that given any starting input the output of next N elements is equally probable (each number and each sequence of N numbers is equally represented in the total period).
the problem statement is the following:
Xorq has invented an encryption algorithm which uses bitwise XOR operations extensively. This encryption algorithm uses a sequence of non-negative integers x1, x2, … xn as key. To implement this algorithm efficiently, Xorq needs to find maximum value for (a xor xj) for given integers a,p and q such that p<=j<=q. Help Xorq to implement this function.
Input
First line of input contains a single integer T (1<=T<=6). T test cases follow.
First line of each test case contains two integers N and Q separated by a single space (1<= N<=100,000; 1<=Q<= 50,000). Next line contains N integers x1, x2, … xn separated by a single space (0<=xi< 2^15). Each of next Q lines describe a query which consists of three integers ai,pi and qi (0<=ai< 2^15, 1<=pi<=qi<= N).
Output
For each query, print the maximum value for (ai xor xj) such that pi<=j<=qi in a single line.
int xArray[100000];
cin >>t;
for(int j =0;j<t;j++)
{
cin>> n >>q;
//int* xArray = (int*)malloc(n*sizeof(int));
int i,a,pi,qi;
for(i=0;i<n;i++)
{
cin>>xArray[i];
}
for(i=0;i<q;i++)
{
cin>>a>>pi>>qi;
int max =0;
for(int it=pi-1;it<qi;it++)
{
int t = xArray[it] ^ a;
if(t>max)
max =t;
}
cout<<max<<"\n" ;
}
No other assumptions may be made except for those stated in the text of the problem (numbers are not sorted).
The code is functional but not fast enough; is reading from stdin really that slow or is there anything else I'm missing?
XOR flips bits. The max result of XOR is 0b11111111.
To get the best result
if 'a' on ith place has 1 then you have to XOR it with key that has ith bit = 0
if 'a' on ith place has 0 then you have to XOR it with key that has ith bit = 1
saying simply, for bit B you need !B
Another obvious thing is that higher order bits are more important than lower order bits.
That is:
if 'a' on highest place has B and you have found a key with highest bit = !B
then ALL keys that have highest bit = !B are worse that this one
This cuts your amount of numbers by half "in average".
How about building a huge binary tree from all the keys and ordering them in the tree by their bits, from MSB to LSB. Then, cutting the A bit-by-bit from MSB to LSB would tell you which left-right branch to take next to get the best result. Of course, that ignores PI/QI limits, but surely would give you the best result since you always pick the best available bit on i-th level.
Now if you annotate the tree nodes with low/high index ranges of its subelements (performed only done once when building the tree), then later when querying against a case A-PI-QI you could use that to filter-out branches that does not fall in the index range.
The point is that if you order the tree levels like the MSB->LSB bit order, then the decision performed at the "upper nodes" could guarantee you that currently you are in the best possible branch, and it would hold even if all the subbranches were the worst:
Being at level 3, the result of
0b111?????
can be then expanded into
0b11100000
0b11100001
0b11100010
and so on, but even if the ????? are expanded poorly, the overall result is still greater than
0b11011111
which would be the best possible result if you even picked the other branch at level 3rd.
I habe absolutely no idea how long would preparing the tree cost, but querying it for an A-PI-QI that have 32 bits seems to be something like 32 times N-comparisons and jumps, certainly faster than iterating randomly 0-100000 times and xor/maxing. And since you have up to 50000 queries, then building such tree can actually be a good investment, since such tree would be build once per keyset.
Now, the best part is that you actually dont need the whole tree. You may build such from i.e. first two or four or eight bits only, and use the index ranges from the nodes to limit your xor-max loop to a smaller part. At worst, you'd end up with the same range as PiQi. At best, it'd be down to one element.
But, looking at the max N keys, I think the whole tree might actually fit in the memory pool and you may get away without any xor-maxing loop.
I've spent some time google-ing this problem and it seams that you can find it in the context of various programming competitions. While the brute force approach is intuitive it does not really solve the challenge as it is too slow.
There are a few contraints in the problem which you need to speculate in order to write a faster algorithm:
the input consists of max 100k numbers, but there are only 32768 (2^15) possible numbers
for each input array there are Q, max 50k, test cases; each test case consists of 3 values, a,pi,and qi. Since 0<=a<2^15 and there are 50k cases, there is a chance the same value will come up again.
I've found 2 ideas for solving the problem: splitting the input in sqrt(N) intervals and building a segment tree ( a nice explanation for these approaches can be found here )
The biggest problem is the fact that for each test case you can have different values for a, and that would make previous results useless, since you need to compute max(a^x[i]), for a small number of test cases. However when Q is large enough and the value a repeats, using previous results can be possible.
I will come back with the actual results once I finish implementing both methods
I need to generate a 10 character unique id (SIP/VOIP folks need to know that it's for a param icid-value in the P-Charging-Vector header). Each character shall be one of the 26 ASCII letters (case sensitive), one of the 10 ASCII digits, or the hyphen-minus.
It MUST be 'globally unique (outside of the machine generating the id)' and sufficiently 'locally unique (within the machine generating the id)', and all that needs to be packed into 10 characters, phew!
Here's my take on it. I'm FIRST encoding the 'MUST' be encoded globally unique local ip address into base-63 (its an unsigned long int that will occupy 1-6 characters after encoding) and then as much as I can of the current time stamp (its a time_t/long long int that will occupy 9-4 characters after encoding depending on how much space the encoded ip address occupies in the first place).
I've also added loop count 'i' to the time stamp to preserve the uniqueness in case the function is called more than once in a second.
Is this good enough to be globally and locally unique or is there another better approach?
Gaurav
#include <stdio.h>
#include <string.h>
#include <sys/time.h>
//base-63 character set
static char set[]="abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789-";
// b63() returns the next vacant location in char array x
int b63(long long longlong,char *x,int index){
if(index > 9)
return index+1;
//printf("index=%d,longlong=%lld,longlong%63=%lld\n",index,longlong,longlong%63);
if(longlong < 63){
x[index] = set[longlong];
return index+1;
}
x[index] = set[longlong%63];
return b63(longlong/63,x,index+1);
}
int main(){
char x[11],y[11] = {0}; /* '\0' is taken care of here */
//let's generate 10 million ids
for(int i=0; i<10000000; i++){
/* add i to timestamp to take care of sub-second function calls,
3770168404(is a sample ip address in n/w byte order) = 84.52.184.224 */
b63((long long)time(NULL)+i,x,b63((long long)3770168404,x,0));
// reverse the char array to get proper base-63 output
for(int j=0,k=9; j<10; j++,k--)
y[j] = x[k];
printf("%s\n",y);
}
return 0;
}
It MUST be 'globally unique (outside
of the machine generating the id)' and
sufficiently 'locally unique (within
the machine generating the id)', and
all that needs to be packed into 10
characters, phew!
Are you in control of all the software generating ids? Are you doling out the ids? If not...
I know nothing about SIP, but there's got to be a misunderstanding that you have about the spec (or the spec must be wrong). If another developer attempts to build an id using a different algorithm than the one you've cooked up, you will have collisions with their ids, meaning they will know longer be globally unique in that system.
I'd go back to the SIP documentation, see if there's an appendix with an algorithm for generating these ids. Or maybe a smarter SO user than I can answer what the SIP algorithm for generating these id's is.
I would have a serious look at RFC 4122 which describes the generation of 128-bit GUIDs. There are several different generation algorithms, some of which may fit (MAC address-based one springs to mind). This is a bigger number-space than yours 2^128 = 3.4 * 10^38 compared with 63^10 = 9.8 * 10^17, so you may have to make some compromises on uniqueness. Consider factors like how frequently the IDs will be generated.
However in the RFC, they have considered some practical issues, like the ability to generate large numbers of unique values efficiently by pre-allocating blocks of IDs.
Can't you just have a distributed ID table ?
Machines on NAT'ed LANs will often have an IP from a small range, and not all of the 32-bit values would be valid (think multicast, etc). Machines may also grab the same timestamp, especially if the granularity is large (such as seconds); keep in mind that the year is very often going to be the same, so it's the lower bits that will give you the most 'uniqueness'.
You may want to take the various values, hash them with a cryptographic hash, and translate that to the characters you are permitted to use, truncating to the 10 characters.
But you're dealing with a value with less than 60 bits; you need to think carefully about the implications of a collision. You might be approaching the problem the wrong way...
Well, if I cast aside the fact that I think this is a bad idea, and concentrate on a solution to your problem, here's what I would do:
You have an id range of 10^63, which correspond to roughly 60 bits. You want it to be both "globally" and "locally" unique. Let's generate the first N bits to be globally unique, and the rest to be locally unique. The concatenation of the two will have the properties you are looking for.
First, the global uniqueness : IP won't work, especially local ones, they hold very little entropy. I would go with MAC addresses, they were made for being globally unique. They cover a range of 256^6, so using up 6*8 = 48 bits.
Now, for the locally unique : why not use the process ID ? I'm making the assumption that the uniqueness is per process, if it's not, you'll have to think of something else. On Linux, process ID is 32 bits. If we wanted to nitpick, the 2 most significant bytes probably hold very little entropy, as they would at 0 on most machines. So discard them if you know what you're doing.
So now you'll see you have a problem as it would use up to 70 bits to generate a decent (but not bulletproof) globally and locally unique ID (using my technique anyway). And since I would also advise to put in a random number (at least 8 bits long) just in case, it definitely won't fit. So if I were you, I would hash the ~78 generated bits to SHA1 (for example), and convert the first 60 bits of the resulting hash to your ID format. To do so, notice that you have a 63 characters range to chose from, so almost the full range of 6 bits. So split the hash in 6 bits pieces, and use the first 10 pieces to select the 10 characters of your ID from the 63 character range. Obviously, the range of 6 bits is 64 possible values (you only want 63), so if you have a 6 bits piece equals to 63, either floor it to 62 or assume modulo 63 and pick 0. It will slightly bias the distribution, but it's not too bad.
So there, that should get you a decent globally and locally pseudo-unique ID.
A few last points: according to the Birthday paradox, you'll get a ~ 1 % chance of collisions after generating ~ 142 million IDs, and a 99% chance after generating 3 billions IDs. So if you hit great commercial success and have millions of IDs being generated, get a larger ID.
Finally, I think I provided a "better than the worse" solution to your problem, but I can't help but think you're attacking this problem in the wrong fashion, and possibly as other have mentioned, misreading the specs. So use this if there are no other ways that would be more "bulletproof" (centralised ID provider, much longer ID ... ).
Edit: I re-read your question, and you say you call this function possibly many times a second. I was assuming this was to serve as some kind of application ID, generated once at the start of your application, and never changed afterwards until it exited. Since it's not the case, you should definitely add a random number and if you generate a lot of IDs, make that at least a 32 bits number. And read and re-read the Birthday Paradox I linked to above. And seed your number generator to a highly entropic value, like the usec value of the current timestamp for example. Or even go so far as to get your random values from /dev/urandom .
Very honestly, my take on your endeavour is that 60 bits is probably not enough...
Hmm, using the system clock may be a weakness... what if someone sets the clock back? You might re-generate the same ID again. But if you are going to use the clock, you might call gettimeofday() instead of time(); at least that way you'll get better resolution than one second.
#Doug T.
No, I'm not in control of all the software generating the ids.
I agree without a standardized algorithm there maybe collisions, I've raised this issue in the appropriate mailing lists.
#Florian
Taking a cue from you're reply. I decided to use the /dev/urandom PRNG for a 32 bit random number as the space unique component of the id. I assume that every machine will have its own noise signature and it can be assumed to be safely globally unique in space at an instant of time. The time unique component that I used earlier remains the same.
These unique ids are generated to collate all the billing information collected from different network functions that independently generated charging information of a particular call during call processing.
Here's the updated code below:
Gaurav
#include <stdio.h>
#include <string.h>
#include <time.h>
//base-63 character set
static char set[]="abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789-";
// b63() returns the next vacant location in char array x
int b63(long long longlong, char *x, int index){
if(index > 9)
return index+1;
if(longlong < 63){
x[index] = set[longlong];
return index+1;
}
x[index] = set[longlong%63];
return b63(longlong/63, x, index+1);
}
int main(){
unsigned int number;
char x[11], y[11] = {0};
FILE *urandom = fopen("/dev/urandom", "r");
if(!urandom)
return -1;
//let's generate a 1 billion ids
for(int i=0; i<1000000000; i++){
fread(&number, 1, sizeof(number), urandom);
// add i to timestamp to take care of sub-second function calls,
b63((long long)time(NULL)+i, x, b63((long long)number, x, 0));
// reverse the char array to get proper base-63 output
for(int j=0, k=9; j<10; j++, k--)
y[j] = x[k];
printf("%s\n", y);
}
if(urandom)
fclose(urandom);
return 0;
}
How can I generate a random number within range 0 to n where n can be > RAND_MAX in c,c++?
Thanks.
split the generation in two phases, then combine the resulting numbers.
Random numbers is a very specialized subject that unless you are a maths junky is very easy to get wrong. So I would advice against building a random number from multiple sources you should use a good library.
I would first look at boost::Random
If that is not suffecient try of this group sci.crypt.random-numbers
Ask the question there they should be able to help.
suppose you want to generate a 64-bit random number, you could do this:
uint64_t n = 0;
for(int i = 0; i < 8; ++i) {
uint64_t x = generate_8bit_random_num();
n = (n << (8 * i)) | x;
}
Of course you could do it 16/32 bits at a time too, but this illustrates the concept.
How you generate that 8/16/32-bit random numbers is up to you. It could be as simple as rand() & 0xff or something better depending on how much you care about the randomness.
Assuming C++, have you tried looking at a decent random number library, like Boost.Random. Otherwise you may have to combine multiple random numbers.
If you're looking for a uniform distribution (or any distribution for that manner) , you must take care that the statistical properties of the output are sufficient for your needs. If you can't use the output of a random number generator directly, you should be very careful trying to combine numbers to achieve your needs.
At a bare minimum you should make sure the distribution is appropriate. If you're looking for a uniform distribution of integers from 0 to M, and you have some uniform random number generator g() to produce outputs that are smaller than M, make sure you do not do one of the following:
add k outputs of g() together until they're large enough (the result is nonuniform)
take r = g() + (g() << 16), then compute r % M (if the range of r is not an even multiple of M, it will weight certain values in the range slightly more than others; the shift-left itself is questionable unless g() outputs a range between 0 and a power of 2 minus 1)
Beyond that, there is the potential for cross-correlation between terms of the sequence (random number generators are supposed to produce independent identically-distributed outputs).
Read The Art of Computer Programming vol. 2 (Knuth) and/or Numerical Recipes and ask questions until you feel confident.
If your implementation has an integer type large enough to hold the result you need, it's generally easier to get a decent distribution by simply using a generator that produces the required range than to try to combine outputs from the smaller generator.
Of course, in most cases, you can just download code for something like the Mersenne Twister or (if you need a cryptographic quality generator) Blum-Blum-Shub, and forget about writing your own.
Do x random numbers (from 0 to RAND_MAX) and add them together, where
x = n % RAND_MAX
Consider a random variable which can take on values {0, 1} with P(0) = P(1) = 0.5. If you want to generate random values between 0 to 2 by summing two independent draws, you will have P(0) = 0.25, P(1) = 0.5 and P(2) = 0.25.
Therefore, use an appropriate library unless you do not care at all about the PDF of the RNG.
See also Chapter 7 in Numerical Recipes. (This is a link to the older edition but that's the one I studied anyway ;-)
There are many ways to do this.
If you are OK with less granularity (higher chance of dupes), then something like (in pseudocode) rand() * n / RAND_MAX will work to spread the values across a larger range. The catch is that in your real code you'll need to avoid overflow, either via casting rand() or n to a large-enough type (e.g. 64-bit int if RAND_MAX is 0xFFFFFFFF) to hold the multiplication result without overflow, or use a multiply-then-divide API (like GNU's MulDiv64 or Win32's MulDiv) which is optimized for this scenario.
If you want granuarity down to each integer, you can call rand() multiple times and append the results. Another answer suggests calling rand() for each 8-bit/16-bit/32-bit chunk depending on size of RAND_MAX.
But, IMHO, the above ideas can rapidly get complicated, inaccurate, or both. Generating random numbers is a solved problem in other libraries, and it's probably much easier to borrow existing code (e.g. from Boost) than try to roll your own. Open source random number generation algorithm in C++? has answers with more links if you want something besides Boost.
[ EDIT: revising after having a busy day... meant to get back and clean up my quick answer this morning, but got pulled away and only getting back now. :-) ]