I have a small program. I have to calculate combination with repetition.
My code:
int factorial(int a){
if (a<0)
return 0;
if (a==0 || a==1)
return 1;
return factorial(a-1)*a;
}
long int combinationWithRepetion(int n, int k){
long int a,b,wyn=0;
wyn=factorial(n+(k-1))/(factorial(k)*factorial(n-1));
return wyn;
}
int main()
{
int k,n=0;
cout<<"Give n: ";
cin>>n;
cout<<"Give k: ";
cin>>k;
cout<<"combination With Repetion for n="<<n<<
" k="<<k<<".\n Is equal to "<<combinationWithRepetion(n,k)<<endl;
return 0;
}
For n=9 and k=6 in Wolfram alfa I get 3003, but in this program the result is 44.
For me the code is fine.
With n=9 and k=6 you compute factorial(14) which is 87,178,291,200 which will overflow a 4-byte int. You need to use something like long long to get an 8-byte int if you want to use this formula.
There are better formulas for computing binomial coefficients which do not rely on computing the full factorials then doing the division. See Binomial coefficient in programming languages, the direct method (rather than using recursion).
In C++ you can use:
int binomial(int N, int K) {
if( K > N - K )
K = N - K;
int c = 1;
for(int i = 0; i < K; ++i) {
c *= (N - i);
c /= (i + 1);
}
return c;
}
So you are calculating (n+k-1) choose k. Substiuting n=9,k=6, it is 14choose6(=3003). But 14! takes more than 36bits to represent, but your int only has 32 bits. A better implementation would be to simplify n!/((n-k)!k!) to n(n-1)...(n-k+1)/k!. Or you can use pascal triangle.
Related
int reverse(int);
int main(){
cout<<"Enter a number : ";
int n{};
cin>>n;
int n_alias{n};
int m=reverse(n);
if (n_alias==m){
cout<<"The number "<<n<<" in reverse is "<<m<<endl;
cout<<"So "<<n<<" a palindrome\n\n";
}
else{
cout<<"The number "<<n<<" in reverse is "<<m<<endl;
cout<<"So "<<n<<" not a palindrome\n\n";
}
}
int reverse(int q){
vector<int> vec{};
int t{};
while (q>0){
t=q%10;
vec.push_back(t);
q-=t;
q/=10;
}
size_t u{vec.size()};
int b{};
for (size_t i{};i<u;i++){
b+=(vec[i])*pow(10,u-i-1);
}
return b;
}
Now 1234321 is working fine, but it is giving wrong result for 121, 12321 and 123454321. I don't know why thats happening when it is using the same algorithm for both test cases.
One more option:
auto rev = [](int n)
{
auto s = std::to_string(n);
return std::stoi(std::string(s.rbegin(), s.rend()));
}
cout << rev(1234) << endl;
Output:
4321
Floating-point arithmetic is not real number arithmetic. Both mantissa and exponent (generally base 2, though base 10 is not unheard of) have a restricted range.
Proving that including floating-point in your integer calculations is accurate is a decidedly non-trivial task, even if you can depend on the floating-point package used in your implementation being completely accurate, and don't allow any shortcuts. Just avoid it.
Anyway, reversing a number can be done simpler:
auto reverse(unsigned long long n) {
constexpr auto base = 10u;
unsigned long long r = 0;
for (; n > 0; n /= base)
r = r * base + n % base;
return r;
}
In my code I am trying to multiply two numbers. The algorithm is simple as (k)*(k-1)^n. I stored the product (k-1)^n in variable p1 and then I multiply it with k. For n=10, k=10 (k-1)^n-1 should be 387420489 and I got this in variable p1 but on multiplying it with k, I get a negative number. I used modulus but instead of 3874208490, I get some other large positive number. What is the correct approach?
#include <iostream>
using namespace std;
typedef long long ll;
ll big = 1000000000 + 7;
ll multiply(ll a, ll b)
{
ll ans = 1;
for (int i = 1; i <= b; i++)
ans = ans * a;
return ans % big;
}
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
ll n, k;
cin >> n >> k;
ll p1 = multiply(k - 1, n - 1);
cout << p1 << endl; // this gives correct value
ll p2 = (k % big) * (p1 % big);
cout << ((p2 + big) % big) % big << endl;
}
}
What is ll type? If it is just int (and I pretty sure it is), it gets overflowed, because 32-bit signed type can't store values more than (2^31)-1, which approximately equals to 2 * 10^9. You can use long long int to make it work, then your code will work with the results less than 2^63.
It's not surprising you get an overflow. I plugged your equation into wolfram alpha, fixing n at 10 and iterating over k from 0 to 100.
The curve gets very vertical, very quickly at around k = 80.
10^21 requires 70 binary bits to represent it, and you only have 63 in a long long.
You're going to have to decide what the limits of this algorithm's parameters are and pick data types corresponding. Perhaps a double would be more suitable?
link to plot is here
So, I am trying to solve the problem: http://codeforces.com/contest/448/problem/D
Bizon the Champion isn't just charming, he also is very smart.
While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the intersection of the i-th row and j-th column equals i·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately. Can you repeat his success?
Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-th number you write out is called the k-th largest number.
Input
The single line contains integers n, m and k (1 ≤ n, m ≤ 5·105; 1 ≤ k ≤ n·m).
Output
Print the k-th largest number in a n × m multiplication table.
What I did was, I applied binary search from 1 to n*m looking for the number which has exactly k elements less than it. For this, I made the following code:
using namespace std;
#define ll long long
#define pb push_back
#define mp make_pair
ll n,m;
int f (int val);
int min (int a, int b);
int main (void)
{
int k;
cin>>n>>m>>k;
int ind = k;
ll low = 1LL;
ll high = n*m;
int ans;
while (low <= high)
{
ll mid = low + (high-low)/2;
if (f(mid) == k)
ans = mid;
else if (f(mid) < k)
low = mid+1;
else
high = mid-1;
}
cout<<ans<<"\n";
return 0;
}
int f (int val)
{
int ret = 0;
for ( int i = 1; i <= n; i++ )
{
ret = ret + min(val/i,m);
}
return ret;
}
int min (int a, int b)
{
if (a < b)
return a;
else
return b;
}
However, I don't know why but this gives wrong answer on test cases:
input
2 2 2
output
2
My output comes out to be 0
I am learning binary search but I don't know where am I going wrong with this implementation. Any help will be appreciated.
Ignoring the fact that your binary search is not the fastest method, you still want to know why it is incorrect.
First be very clear about what you want and what your f is returning:
looking for the number which has exactly k elements less than it.
No! You are looking for the smallest number that has k elements less than or equal to it. And your function f(X) returns the count of elements less than or equal to X.
So when f(X) returns a value too small, you know X must be larger by at least 1, so low=mid+1 is correct. But when f(X) returns a value too large, X might be perfect (might be an element appearing several times in the table). Conversely, when f(X) returns exactly the right number, X might still be too big (X might be a value that appears zero times in the table).
So when f(X) is not too small, the best you can do is high=mid not high=mid-1
while (low < high)
{
ll mid = low + (high-low)/2;
if (f(mid) < k)
low = mid+1;
else
high = mid;
}
Notice low never gets > high, so stop when they are equal, and we don't try to catch the ans along the way. Instead at the end low==high==Answer
The contest says 1 second time limit. On my computer, your code with that correction solves the max size problem in under a second. But I'm not sure the judging computer is that fast.
Edit: int is too small for the max size of the problem, so you can't return int from f:
n, m, and i each fit in 32 bits, but the input and output of f() as well as k, ret, low and high all need to hold integers up to 2.5e11
import java.util.*;
public class op {
static int n,m;
static long k;
public static void main(String args[]){
Scanner s=new Scanner(System.in);
n=s.nextInt();
m=s.nextInt();
k=s.nextLong();
long start=1;
long end=n*m;
long ans=0;
while(end>=start){
long mid=start+end;
mid/=2;
long fmid=f(mid);
long gmid=g(mid);
if(fmid>=k && fmid-gmid<k){
ans=mid;
break;
}
else if(f(mid)>k){
end=mid-1;
}
else{
start=mid+1;
}
}
System.out.println(ans);
}
static long f (long val)
{
long ret = 0;
for ( int i = 1; i <= n; i++ )
{
ret = ret + Math.min(val/i,m);
}
return ret;
}
static long g (long val)
{
long ret = 0;
for ( int i = 1; i <= n; i++ )
{
if(val%i==0 && val/i<=m){
ret++;
}
}
return ret;
}
public static class Pair{
int x,y;
Pair(int a,int b){
x=a;y=b;
}
}
}
Find the smallest number which is divisible by all numbers from 1 to N, without leaving any remainder. As number can be very large we take the answer modulo 1000000007.
I think the smallest number that would be divisible by all the number from 1 to N,would be LCM(1..N).
Example: for N = 5, that smallest number would be 60.
As 60 is the smallest number divisible by all the number form (1-5).
But for some strange reason its giving me wrong answer for large N(1000), etc.
What can cause the possible error here, is my login correct here?
Here's what i tried to Implement.
#include <iostream>
#include <vector>
using namespace std;
vector<long long> lcmArr;
const long long mod = 1000000007;
long long gcd(long long a, long long b){
if(b == 0)
{
return a;
}
return gcd(b, a%b);
}
long long lcmFumction(long long a, long long b)
{
return (a*b)/gcd(a,b);
}
int main() {
lcmArr.clear();lcmArr.resize(1002);
lcmArr[0] =0; lcmArr[1] = 1;
for(int i =2; i <=1000; i++){
lcmArr[i] = lcmFumction(lcmArr[i-1], i)%mod;
//cout<<lcmArr[i-1]<<" ";
}
int T;
cin >> T;
while(T--) {
int N;
cin>>N;
cout<<lcmArr[N]<<"\n";
}
return 0;
}
The problem is when you calculate LCM, you use division,
And
((A/B)/C) mod M != (((A/B) mod M)/C)mod M
For example (10/5/2) % 2 != ((10/5)%2)/2)%2
You should use modular inverse to calculate that.
Some explanation about modular inverse.
If we have:
(a*b) % m = 1, then b is modular inverse of a, as b % m = (1/a) % m.
Thus, if we need to calculate (x/a) % m, we can make it become (x * b ) %m.
And we know that (A*B*C)% m = ((A * B) % m)*C)% m, so, in your case, modular inverse will come in handy.
I know the answer above has already been accepted, but I think that won't be enough to solve your problem. The problem lies in the fact that the first modular LCM will take away all divisors you need to check in subsequent GCD calls, so the answer will still be wrong.
One possible solution is to keep an array of factors for the answer. Each factor will be each number from 1..N, divided by GCD(number, [all previous numbers]). For this to work, you have to code a special GCD that computes the result between a single number and an array of factors. This C++ code shows how this would work:
#include <iostream>
#include <vector>
#define lli long long int
using namespace std;
vector<lli> T;
lli gcd(lli a, lli b) {
if(b == 0)
return a;
return gcd(b, a%b);
}
lli gcd_vector(vector<lli>& a, lli b) {
lli ma = 1;
for(int i=0; i<T.size(); i++)
ma = ma*T[i]%b;
return gcd(b, ma);
}
int main() {
lli answer = 1;
for(int i=1; i<=1000; i++) {
lli factor = i/gcd_vector(T, i);
T.push_back(factor);
answer = (answer*factor)%1000000007;
}
cout << answer << endl;
}
I am doing a factorial program with strings because i need the factorial of Numbers greater than 250
I intent with:
string factorial(int n){
string fact="1";
for(int i=2; i<=n; i++){
b=atoi(fact)*n;
}
}
But the problem is that atoi not works. How can i convert my string in a integer.
And The most important Do I want to know if the program of this way will work with the factorial of 400 for example?
Not sure why you are trying to use string. Probably to save some space by not using integer vector? This is my solution by using integer vector to store factorial and print.Works well with 400 or any large number for that matter!
//Factorial of a big number
#include<iostream>
#include<vector>
using namespace std;
int main(){
int num;
cout<<"Enter the number :";
cin>>num;
vector<int> res;
res.push_back(1);
int carry=0;
for(int i=2;i<=num;i++){
for(int j=0;j<res.size();j++){
int tmp=res[j]*i;
res[j]=(tmp+carry)%10 ;
carry=(tmp+carry)/10;
}
while(carry!=0){
res.push_back(carry%10);
carry=carry/10;
}
}
for(int i=res.size()-1;i>=0;i--) cout<<res[i];
cout<<endl;
return 0;
}
Enter the number :400
Factorial of 400 :64034522846623895262347970319503005850702583026002959458684445942802397169186831436278478647463264676294350575035856810848298162883517435228961988646802997937341654150838162426461942352307046244325015114448670890662773914918117331955996440709549671345290477020322434911210797593280795101545372667251627877890009349763765710326350331533965349868386831339352024373788157786791506311858702618270169819740062983025308591298346162272304558339520759611505302236086810433297255194852674432232438669948422404232599805551610635942376961399231917134063858996537970147827206606320217379472010321356624613809077942304597360699567595836096158715129913822286578579549361617654480453222007825818400848436415591229454275384803558374518022675900061399560145595206127211192918105032491008000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
There's a web site that will calculate factorials for you: http://www.nitrxgen.net/factorialcalc.php. It reports:
The resulting factorial of 250! is 493 digits long.
The result also contains 62 trailing zeroes (which constitutes to 12.58% of the whole number)
3232856260909107732320814552024368470994843717673780666747942427112823747555111209488817915371028199450928507353189432926730931712808990822791030279071281921676527240189264733218041186261006832925365133678939089569935713530175040513178760077247933065402339006164825552248819436572586057399222641254832982204849137721776650641276858807153128978777672951913990844377478702589172973255150283241787320658188482062478582659808848825548800000000000000000000000000000000000000000000000000000000000000
Many systems using C++ double only work up to 1E+308 or thereabouts; the value of 250! is too large to store in such numbers.
Consequently, you'll need to use some sort of multi-precision arithmetic library, either of your own devising using C++ string values, or using some other widely-used multi-precision library (GNU GMP for example).
The code below uses unsigned double long to calculate very large digits.
#include<iostream.h>
int main()
{
long k=1;
while(k!=0)
{
cout<<"\nLarge Factorial Calculator\n\n";
cout<<"Enter a number be calculated:";
cin>>k;
if (k<=33)
{
unsigned double long fact=1;
fact=1;
for(int b=k;b>=1;b--)
{
fact=fact*b;
}
cout<<"\nThe factorial of "<<k<<" is "<<fact<<"\n";
}
else
{
int numArr[10000];
int total,rem=0,count;
register int i;
//int i;
for(i=0;i<10000;i++)
numArr[i]=0;
numArr[10000]=1;
for(count=2;count<=k;count++)
{
while(i>0)
{
total=numArr[i]*count+rem;
rem=0;
if(total>9)
{
numArr[i]=total%10;
rem=total/10;
}
else
{
numArr[i]=total;
}
i--;
}
rem=0;
total=0;
i=10000;
}
cout<<"The factorial of "<<k<<" is \n\n";
for(i=0;i<10000;i++)
{
if(numArr[i]!=0 || count==1)
{
cout<<numArr[i];
count=1;
}
}
cout<<endl;
}
cout<<"\n\n";
}//while
return 0;
}
Output:
![Large Factorial Calculator
Enter a number be calculated:250
The factorial of 250 is
32328562609091077323208145520243684709948437176737806667479424271128237475551112
09488817915371028199450928507353189432926730931712808990822791030279071281921676
52724018926473321804118626100683292536513367893908956993571353017504051317876007
72479330654023390061648255522488194365725860573992226412548329822048491377217766
50641276858807153128978777672951913990844377478702589172973255150283241787320658
18848206247858265980884882554880000000000000000000000000000000000000000000000000
000000000000][1]
You can make atoi compile by adding c_str(), but it will be a long way to go till getting factorial. Currently you have no b around. And if you had, you still multiply int by int. So even if you eventually convert that to string before return, your range is still limited. Until you start to actually do multiplication with ASCII or use a bignum library there's no point to have string around.
Your factorial depends on conversion to int, which will overflow pretty fast, so you want be able to compute large factorials that way. To properly implement computation on big numbers you need to implement logic as for computation on paper, rules that you were tought in primary school, but treat long long ints as "atoms", not individual digits. And don't do it on strings, it would be painfully slow and full of nasty conversions
If you are going to solve factorial for numbers larger than around 12, you need a different approach than using atoi, since that just gives you a 32-bit integer, and no matter what you do, you are not going to get more than 2 billion (give or take) out of that. Even if you double the size of the number, you'll only get to about 20 or 21.
It's not that hard (relatively speaking) to write a string multiplication routine that takes a small(ish) number and multiplies each digit and ripples the results through to the the number (start from the back of the number, and fill it up).
Here's my obfuscated code - it is intentionally written such that you can't just take it and hand in as school homework, but it appears to work (matches the number in Jonathan Leffler's answer), and works up to (at least) 20000! [subject to enough memory].
std::string operator*(const std::string &s, int x)
{
int l = (int)s.length();
std::string r;
r.resize(l);
std::fill(r.begin(), r.end(), '0');
int b = 0;
int e = ~b;
const int c = 10;
for(int i = l+e; i != e;)
{
int d = (s[i]-0x30) * x, p = i + b;
while (d && p > e)
{
int t = r[p] - 0x30 + (d % c);
r[p] = (t % c) + 0x30;
d = t / c + d / c;
p--;
}
while (d)
{
r = static_cast<char>((d % c) +0x30)+r;
d /= c;
b++;
}
i--;
}
return r;
}
In C++, the largest integer type is 'long long', and it hold 64 bits of memory, so obviously you can't store 250! in an integer type. It is a clever idea to use strings, but what you are basically doing with your code is (I have never used the atoi() function, so I don't know if it even works with strings larger than 1 character, but it doesn't matter):
covert the string to integer (a string that if this code worked well, in one moment contains the value of 249!)
multiply the value of the string
So, after you are done multiplying, you don't even convert the integer back to string. And even if you did that, at one moment when you convert the string back to an integer, your program will crash, because the integer won't be able to hold the value of the string.
My suggestion is, to use some class for big integers. Unfortunately, there isn't one available in C++, so you'll have to code it by yourself or find one on the internet. But, don't worry, even if you code it by yourself, if you think a little, you'll see it's not that hard. You can even use your idea with the strings, which, even tough is not the best approach, for this problem, will still yield the results in the desired time not using too much memory.
This is a typical high precision problem.
You can use an array of unsigned long long instead of string.
like this:
struct node
{
unsigned long long digit[100000];
}
It should be faster than string.
But You still can use string unless you are urgent.
It may take you a few days to calculate 10000!.
I like use string because it is easy to write.
#include <bits/stdc++.h>
#pragma GCC optimize (2)
using namespace std;
const int MAXN = 90;
int n, m;
int a[MAXN];
string base[MAXN], f[MAXN][MAXN];
string sum, ans;
template <typename _T>
void Swap(_T &a, _T &b)
{
_T temp;
temp = a;
a = b;
b = temp;
}
string operator + (string s1, string s2)
{
string ret;
int digit, up = 0;
int len1 = s1.length(), len2 = s2.length();
if (len1 < len2) Swap(s1, s2), Swap(len1, len2);
while(len2 < len1) s2 = '0' + s2, len2++;
for (int i = len1 - 1; i >= 0; i--)
{
digit = s1[i] + s2[i] - '0' - '0' + up; up = 0;
if (digit >= 10) up = digit / 10, digit %= 10;
ret = char(digit + '0') + ret;
}
if (up) ret = char(up + '0') + ret;
return ret;
}
string operator * (string str, int p)
{
string ret = "0", f; int digit, mul;
int len = str.length();
for (int i = len - 1; i >= 0; i--)
{
f = "";
digit = str[i] - '0';
mul = p * digit;
while(mul)
{
digit = mul % 10 , mul /= 10;
f = char(digit + '0') + f;
}
for (int j = 1; j < len - i; j++) f = f + '0';
ret = ret + f;
}
return ret;
}
int main()
{
freopen("factorial.out", "w", stdout);
string ans = "1";
for (int i = 1; i <= 5000; i++)
{
ans = ans * i;
cout << i << "! = " << ans << endl;
}
return 0;
}
Actually, I know where the problem raised At the point where we multiply , there is the actual problem ,when numbers get multiplied and get bigger and bigger.
this code is tested and is giving the correct result.
#include <bits/stdc++.h>
using namespace std;
#define mod 72057594037927936 // 2^56 (17 digits)
// #define mod 18446744073709551616 // 2^64 (20 digits) Not supported
long long int prod_uint64(long long int x, long long int y)
{
return x * y % mod;
}
int main()
{
long long int n=14, s = 1;
while (n != 1)
{
s = prod_uint64(s , n) ;
n--;
}
}
Expexted output for 14! = 87178291200
The logic should be:
unsigned int factorial(int n)
{
unsigned int b=1;
for(int i=2; i<=n; i++){
b=b*n;
}
return b;
}
However b may get overflowed. So you may use a bigger integral type.
Or you can use float type which is inaccurate but can hold much bigger numbers.
But it seems none of the built-in types are big enough.