So I am a beginner in c++ programming and I was doing some problem online.
I have to calculate all the products of numbers from 999 to 100 (Eg.999*999 , 999*998 ... 800* 800 , 800 *799 ... , 100 * 100). I can easily print out these products but when I try to pass these values to a function they do not work.
Can you please look at the following code and point out anything that's wrong?
I think its got something to do with buffer but I have no idea how to fix that. Thanks.
#include <iostream>
using namespace std;
unsigned long int num,rev,temp,rem = 0,reversed = 0;
int ispalin(unsigned long int n)
{
temp=n;
while(temp!=0)
{
rem = temp%10;
reversed = reversed*10 + rem;
temp/=10;
}
if(reversed == n)
{
return 1;
}
return 0;
}
int main()
{
int maxi = 0;
for (int i =999 ; i >= 100;i--)
{
for(int j = i;j >= 100; j--)
{
rev = ispalin(i*j);
if (rev == 1)
{
if(i*j > maxi)
{
maxi = i*j;
}
}
}
}
cout<<maxi<<" This is max"<<endl;
}
reversed must be reset to zero at the beginning of every check for palindrome. The best would be to make reversed (and others) a local variable of ispalin.
Related
Background:
This problem comes from leetcode.com
Write an algorithm to determine if a number is "happy".
A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.
Example: 19 is a happy number
1^2 + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1
Question:
I thought of doing a recursion for this particular problem to keep repeating the squaring of the integers until we arrive at 1. I am new with recursion (just read Absolute C++ Ch 13 --- Recursion yesterday).I thought I would give this problem a shot but I am having some trouble.
When I call my function I created I should get a return of 19 since 19 is a "Happy Number", but instead my function just returns 0, and I am not sure why. I just need some help with my approach I have taken and suggestions to changes in my code.
Here is my code:
#include <algorithm>
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
int Happy(int n) {
vector<int> nums;
int length = to_string(n).length();
for(int i = 0; i < length; i++) {
int digit = n % 10;
n /= 10;
nums.push_back(digit);
}
reverse(nums.begin(), nums.end());
int sum = 0;
for(int i = 0; i < length; i++) {
sum += pow(nums[i],2);
}
if (sum == 1) {
return n;
}
else {
return Happy(sum);
}
}
int main() {
int n = 19;
int result = Happy(n);
cout << result << endl;
return 0;
}
Again, I am not sure why I get 0 as the result, when it should return 19.
You forgot to place a return in your code, Also you n becomes 0, and you are returning n when you find sum == 1. It should return the original_num.
To Store the original number reference pass it along with your call to happy method.
#include <algorithm>
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
int Happy(int n, int original_num) {
vector<int> nums;
int length = to_string(n).length();
for(int i = 0; i < length; i++) {
int digit = n % 10;
n /= 10;
nums.push_back(digit);
}
//reverse(nums.begin(), nums.end());
int sum = 0;
for(int i = 0; i < length; i++) {
sum += nums[i]*nums[i];
}
if (sum == 1) {
return original_num;
}
else {
return Happy(sum, original_num);
}
}
int main() {
int n = 19;
int result = Happy(n, n);
cout << result << endl;
return 0;
}
Hope this helps!
I have to find multiplicity of smallest prime factor in all numbers till 10^7.I am using Sieve of Eratosthenes to find all the prime numbers. And there in a seperate array phi i am storing smallest prime factors of composite numbers.Here is my code for that
for(ull i=2;i<=m;i++)
{
if (check[i])
{
uncheck[i]=true;
for (ull k=i*i; k<=n; k+=i)
{
if(check[k]==true)
phi[k]=g;
check[k]=false;
}
}
}
Now i am running a loop till n and using a loop inside it to calculate it.
Here is code for that
for(ull i=4;i<=n;i++)
{
if(check[i]==false)
{
ull count=0;
ull l=i;
ull r=phi[i];
while(l%r==0)
{
l=l/r;
count++;
}
cout<<count<<'\n';
}
}
Is there any faster way to compute this?
Absolutely, you can do this without a loop.
c is probably at most 64 bits. It cannot contain any factor other than 1 more than 63 times. So instead of a loop, you write 63 nested if-statements.
For the case j == 2 your compiler may have some intrinsic functions that count trailing zero bits. If that is the case, then you handle that case separately and you need only 40 if's, because 3^41 > 2^64.
If you want to evaluate n such that jn = c, then recast the problem to
n = log(c) / log(j).
If n is an integer then your problem is solved.
Of course you need to consider floating point precision here; n might not be an exact integer, but close to one.
One alternative option, though not necessarily the most efficient, is to write a simple recursive function, such as this, assuming you are dealing with ints:
int recurseSubtract(int c, int j, int count){
if ((c==j)) {
return count + 1;
} else {
c = c-j;
subtract(c, j, count++);
}
}
int count = recurseSubtract(c,j,0);
However, see here for the pros and cons of loops vs. recursion.
Since you asked for the "multiplicity of smallest prime factor" you could easily use the same sieve approach to get multiplicity as you used to get the smallest factor.
for(ull i=2;i<=m;i++)
{
if (check[i])
{
uncheck[i]=true; // WHY??
ull k=i*i;
for (ull q=i; q<maxq; k=(q*=i))
for ( ; k<=n; k+=q)
{
if(check[k]==true)
phi[k]=g; // I copied 'g' from you, but didn't you mean 'i'?
if ( phi[k]==g )
count[k]++;
check[k]=false;
}
}
}
If you want to do a little better than that, the step of phi[k]==g and the some of the redundancy in check[k] access are needed only because q values are processed in forward sequence. It would be faster to work with q in reverse. Since q's are only easily computed in forward sequence and there are fairly few q's per i, the easiest way to process q backward would be to convert the loop over q into a recursive function (compute q on the way in and process it after the recursive call).
I found one simple rule but can not really describe in words. Here is another code calculating primenumbers
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
double f_power(double val, int exp);
int main(int argc,char* argv[]) {
int p[2];
int ctr = 0;
int ctr2 = 0;
int it_m = 0;
int it_1 = 0;
int it_2 = 0;
int it_c = 0;
int index = 3;
srand(time(NULL));
double t = clock();
double s = clock();
int prime = 2;
FILE *file;
file = fopen("ly_prime.txt", "w");
//f_power(2.0, 57885161)
for (it_m = 2; it_m <= 2000; it_m++) {
for (it_1 = it_m, ctr2 = 0, it_c = it_m; it_1 >= 2; it_1--) {
for (it_2 = it_1; it_2 >= 2; it_2--) {
if (it_1 * it_2 - it_c == 0) {
p[ctr % 2] = it_c;
if (ctr >= 1 && p[ctr % 2] - p[(ctr - 1) % 2] == 2) {
//prime[0] = (p[ctr % 2] - 1);
prime = (p[ctr % 2] - 1);
fprintf(stdout, "|%d _ i: %d _ %d\n", isPrime(prime),index, prime);
index++;
}
ctr++;
}
}
}
}
t = clock() - t;
fprintf(file, "|%d_ %d_ %d ", prime, index - 2, ctr);
}
double f_power(double val, int exp) {
int i = 0;
double help = val;
for(i = 1; i < exp; i++) {
val *= help;
}
return val;
}
int isPrime(int number)
{
int i = 2;
for(i=2; i < number; i++)
{
int leftOver=(number % i);
if (leftOver==0)
{
return 1;
break;
}
}
return 0;
}
perhaps it helps understanding, best regards
The problem from uva OJ
my solution with recursion
#include <cstdio>
using namespace std;
#define garbaze 0
//number of ways changes can be made
int coins[] = {garbaze,50,25,10,5,1}; //order does not matter//as in the //count_ways... function we are returning
//0 if which_coin_now is <= 0 so it
//does n't matter what we have in the index 0 [garbaze] .. but we must put //something there to implement the
//code using the pseudo code or recursive relation
typedef unsigned long long ull; //simple typedef
ull dp[7490][6]; //2d table
//recursive approach
ull count_ways_of_changes(int money_now,int which_coin_now)
{
if(money_now == 0)
return 1;
if(money_now < 0 || which_coin_now <=0 )
return 0;
if(dp[money_now][which_coin_now] == -1)
dp[money_now][which_coin_now] = count_ways_of_changes(money_now,which_coin_now-1) //excluding current coin
+ count_ways_of_changes(money_now - coins[which_coin_now],which_coin_now) ; //including current coin
return dp[money_now][which_coin_now] ;
}
int main()
{
for(int loop = 0; loop< 7490 ;loop++)
for(int sec_loop = 0;sec_loop<6;sec_loop++)
dp[loop][sec_loop] = -1; //table initialization
int N = 0;
while(scanf("%d",&N)==1)
{
printf("%llu\n",count_ways_of_changes(N,5)); //llu for unsigned long long
}
return 0;
}
This one got accepted (and took 0.024 s)
And my iterative approach :
#include <cstdio>
//#include <iostream>
//using namespace std;
typedef unsigned long long ull;
ull dp[7490][6];
#define garbaze 0
int value_coins[] = {garbaze,5,1,10,25,50} ;
inline ull count_ways_change(int money,int num_of_coins)
{
for(int sum_money_now = 0; sum_money_now <= money ;sum_money_now++)
for(int recent_coin_index = 0 ; recent_coin_index <= num_of_coins ; recent_coin_index++)
//common mistakes : starting the second index at num_of_coins and decrementing till 0 ...see we are pre calculating
//we have to start bottom to up....if we start at dp[0][5] .....to dp[1][5] but to know that i need to know
//dp[1][4] and dp[..][5] before hand ..but we have not calculated dp[1][4] yet...in this case i don't go to infinite
//loop or anything as the loop is well defined but i get stupid garbaze answer
{
if(sum_money_now == 0)
dp[sum_money_now][recent_coin_index] = 1;
else if(recent_coin_index == 0)
dp[sum_money_now][recent_coin_index] = 0;
else if(sum_money_now < value_coins[recent_coin_index] && recent_coin_index != 0)
dp[sum_money_now][recent_coin_index] = dp[sum_money_now][recent_coin_index-1] ;
else
dp[sum_money_now][recent_coin_index] = dp[sum_money_now][recent_coin_index-1] + dp[sum_money_now - value_coins[recent_coin_index] ][recent_coin_index] ;
// cout<<dp[sum_money_now][recent_coin_index]<<endl;
}
return dp[money][num_of_coins] ;
}
int main()
{/*
for(int loop = 0; loop< 7490 ;loop++)
for(int sec_loop = 0;sec_loop<6;sec_loop++)
dp[loop][sec_loop] = -1; //table initialization
*/ //In the iterative version do not need to initialize the table as we are working bottom - up
int N = 0;
while(scanf("%d",&N)==1)
{
printf("%llu\n",count_ways_change(N,5)); //llu for unsigned long long
}
return 0;
}
But i got time limit exceeded for this one.It gives correct output but i don't see a reason why this one has to be so slow?
The difference is your recursive solution remember partial solutions from previous tasks (because the DP table is global and does not get removed between different inputs), while the iterative doesn't - for each new input, it recalculates the DP matrix from scratch.
It can be solved by remembering which portion of the DP table was already calculated and avoid recalculating it, rather than recalculate it from scratch for every query.
I'm trying to solve a problem from Codeforces (http://codeforces.com/problemset/problem/189/A)
Here's the problem statement:
Polycarpus has a ribbon, its length is n. He wants to cut the ribbon in a way that fulfils the following two conditions:
After the cutting each ribbon piece should have length a, b or c.
After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required
cutting.
Input
The first line contains four space-separated integers n, a, b and c (1 ≤ n, a, b, c ≤ 4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers a, b and c can coincide.
Output
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.
Sample Input
5 5 3 2
Sample Output
2
I tried to solve this problem using Dynamic Programming (Topdown approach). But I'm not able to get the correct answer. There might be something wrong with the recursive function. Here's my code:
#include<bits/stdc++.h>
using namespace std;
int n,s;
int a[3];
int val,m=-1;
int dp(int n)
{
if(n==0)
return 0;
for(int i=0;i<3;i++)
{
if(n>=a[i])
{
val=1+dp(n-a[i]);
}
}
if(val>m)
m=val;
return m;
}
int main()
{
scanf("%d %d %d %d",&n,&a[0],&a[1],&a[2]);
cout<<dp(n)<<endl;
return 0;
}
What is the problem in the above approach?
There are several problems:
Wrong Search
In your lines
for(int i=0;i<3;i++)
{
if(n>=a[i])
{
val=1+dp(n-a[i]);
}
}
if(val>m)
m=val;
You should be checking for the maximum of the different vals obtained for the different choices of i.
Wrong Termination
If the length is not 0 and no ribbon can be cut, you should return something like minus infinity. You currently return m which is initially -1 (more on this later). This is wrong, and for long ribbons will essentially ensure that you just choose the minimum of a, b, and c.
Use of Globals
Some globals, e.g., m are initialized once but are modified by the recursion. It's not "just" bad programming habits - it's not doing what you want.
No Reuse
By calling the recursion unconditionally, and not reusing previous calls, your running time is needlessly high.
int main() {
int n, a, b, c;
scanf("%d %d %d %d", &n, &cuts[0], &cuts[1], &cuts[2]);
sort(cuts, cuts + 3);
for (int i = 0; i <= n; i++) {
max_cuts[i] = INT_MIN;
}
max_cuts[0] = 0;
max_cuts[cuts[0]] = 1;
max_cuts[cuts[1]] = 1;
max_cuts[cuts[2]] = 1;
for (int i = 1; i <= n; i++) {
for (int j = 0; j < 3; j++) {
if (cuts[j] > i) break;
max_cuts[i] = max(max_cuts[i - cuts[j]] + 1, max_cuts[i]);
}
}
printf("%d\n", max_cuts[n]);
return 0;
}
#Ami Tavory correctly suggested the problems with your recursive approach. May be my solution below can help you understand better how to form states and check bounds:
int main()
{
int n, a, b, c;
cin >> n >> a >> b >> c;
const int l = n + 1;
int sum[l];
fill(sum, sum+l, INT_MIN);
sum[0] = 0;
for(int i=1; i<=n; i++)
{
if(i - a >= 0)
{
sum[i] = sum[i-a] + 1;
}
if(i - b >= 0 && sum[i-b] + 1 > sum[i])
{
sum[i] = sum[i-b] + 1;
}
if(i - c >= 0 && sum[i-c] + 1 > sum[i])
{
sum[i] = sum[i-c] + 1;
}
}
cout << sum[n] << endl;
return 0;
}
Simply at each sum[i], we are maximizing the number of cuts. So, at sum[i], we are storing the max(sum[i-a]+1, sum[i-b]+1, sum[i-c]+1).
Other than this, there are just bound checks.
you can solve this problem through top down approach.A dp problem always check all the possible cases then gives us the optimal solution.so here is the code
#include<bits/stdc++.h>
using namespace std;
int a,b,c;
int DP[4001];
int solve(int n){
if(n == 0)return 0;
if(n<0) return INT_MIN;
if(DP[n] != -1)return DP[n];
else{
DP[n] = max(1+solve(n-a),max(1+solve(n-b),1+solve(n-c)));
return DP[n];
}
}
int main(){
int n,x;
cin>>n>>a>>b>>c;
for(int i = 0;i<4001;++i){
DP[i] = -1;
}
x = solve(n);
cout<<x;
}
Here is my implementation of Problem 25 - Project Euler (see comments in code for explanation of how it works):
#include <iostream> //Declare headers and use correct namespace
#include <math.h>
using namespace std;
//Variables for the equation F_n(newTerm) = F_n-1(prevTerm) + Fn_2(currentTerm)
unsigned long long newTerm = 0;
unsigned long long prevTerm = 1; //F_1 initially = 1
unsigned long long currentTerm = 1; //F_2 initially = 2
unsigned long long termNo = 2; //Current number for the term
void getNextTerms() { //Iterates through the Fib sequence, by changing the global variables.
newTerm = prevTerm + currentTerm; //First run: newTerm = 2
unsigned long long temp = currentTerm; //temp = 1
currentTerm = newTerm; //currentTerm = 2
prevTerm = temp; //prevTerm = 1
termNo++; //termNo = 3
}
unsigned long long getLength(unsigned long long number) //Returns the length of the number
{
unsigned long long length = 0;
while (number >= 1) {
number = number / 10;
length++;
}
return length;
}
int main (int argc, const char * argv[])
{
while (true) {
getNextTerms(); //Gets next term in the Fib sequence
if (getLength(currentTerm) < 1000) { //Checks if the next terms size is less than the desired length
}
else { //Otherwise if it is perfect print out the term.
cout << termNo;
break;
}
}
}
This works for the example, and will run quickly as long as this line:
if (getLength(currentTerm) < 1000) { //Checks if the next term's size is less than the desired length
says 20 or lower instead of 1000. But if that number is greater than 20 it takes a forever, my patience gets the better of me and I stop the program, how can I make this algorithm more efficient?
If you have any questions just ask in the comments.
There is a closed formula for the Fibonachi numbers (as well as for any linear recurrent sequence).
So F_n = C1 * a^n + C2 * b^n, where C1, C2, a and b are numbers that can be found from the initial conditions, i.e. for the Fib case from
F_n+2 = F_n+1 + F_n
F_1 = 1
F_2 = 1
I don't give their values on purpose here. It's just a hint.
nth fibonacci number is =
(g1^n-g2^n)/sqrt(5).
where g1 = (1+sqrt(5))/2 = 1.61803399
g2 = (1-sqrt(5))/2 = -0.61803399
For finding the length of nth fibonacci number, we can just calculate the log(nth fibonacci number).So, length of nth fibonacci number is,
log((g1^n-g2^n)/sqrt(5)) = log(g1^n-g2^n)-0.5*log(5).
you can just ignore g2^n, since it is very small negative number.
Hence, length of nth fibonacci is
n*log(g1)-0.5*log(5)
and we need to find the smallest value of 'n' such that this length = 1000, so we can find the value of n for which the length is just greater than 999.
So,
n*log(g1)-0.5*log(5) > 999
n*log(g1) > 999+0.5*log(5)
n > (999+0.5*log(5))/log(g1)
n > (999.3494850021680094)/(0.20898764058551)
n > 4781.859263075
Hence, the smallest required n is 4782. No use of any coding, easiest way.
Note: everywhere log is used in base 10.
This will probably speed it up a fair bit:
int getLength(unsigned long long number) //Returns the length of the number when expressed in base-10
{
return (int)log10(number) + 1;
}
...but, you can't reach 1000 digits using an unsigned long long. I suggest looking into arbitrary-precision arithmetic libraries, or languages which have arbitrary-precision arithmetic built in.
You could try computing a Fibonacci number using matrix exponentiation. Then repeated doubling to get to a number that has more than 1000 digits and use binary search in that range to find the first one.
using doubles, you can come to a solution knowing the highest exponential is 308:
get the sequence to the exp of 250, then divide your two numbers by 1e250. Restart the algorithm with those two numbers
if you do this 4 times, you'll get the right answer
C++ code maybe as follows:
#include "iostream"
#include "string.h"
#include "algorithm"
using namespace std;
string addTwoString(string a, string b)
{
if (a.length() == 0)
{
return b;
}
if (b.length() == 0)
{
return a;
}
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
string result = "";
string str_1, str_2;
if (a.length() > b.length())
{
str_1 = b;
str_2 = a;
}
else
{
str_1 = a;
str_2 = b;
}
int index = 0;
int value = 0, over_value = 0;
for (; index < str_1.length(); ++index)
{
int temp_1 = (int)(str_1[index] - '0');
int temp_2 = (int)(str_2[index] - '0');
int temp = temp_1 + temp_2 + over_value;
value = temp % 10;
over_value = temp / 10;
char c = (char)(value + '0');
result += c;
}
for (; index < str_2.length(); ++index)
{
int temp_2 = (int)(str_2[index] - '0');
int temp = temp_2 + over_value;
value = temp % 10;
over_value = temp / 10;
char c = (char)(value + '0');
result += c;
}
if (over_value > 0)
{
char c = (char)(over_value + '0');
result += c;
}
reverse(result.begin(), result.end());
return result;
}
int main()
{
string a = "1";
string b = "1";
string c = addTwoString(a, b);
int index = 3;
while (c.length() < 1000)
{
a = b;
b = c;
c = addTwoString(a, b);
++ index;
}
cout << index << endl;
}
I just used a recursive function that adds arrays vertically to complete the problem. Basically zero run time, less than 50 lines of code. Enjoy:
#include <stdio.h>
int Calc_Fib (int numA[], int numB[], int temp[], int index) {
int i = 0;
//Check 1000th digit for non-zero value.
if (numB[999] != 0) return index;
//Add arrays A and B vertically.
for (i = 0; i < 1000; ++i) {
temp[i] += (numA[i] + numB[i]);
if (temp[i] > 9) {
temp[i + 1] = temp[i] / 10;
temp[i] %= 10;
}
numA[i] = numB[i];
numB[i] = temp[i];
temp[i] = 0;
}
Calc_Fib(numA, numB, temp, ++index);
}
int main() {
int numA[1000]; //Holds previous term.
int numB[1000]; //Holds current term.
int temp[1000]; //Holds temporary number for vertical addition.
int i = 0;
int indexVal = 2;
for (i = 0; i < 1000; ++i) {
numA[i] = 0;
numB[i] = 0;
temp[i] = 0;
}
//Initialize first two terms.
numA[0] = (numB[0] = 1);
indexVal = Calc_Fib(numA, numB, temp, indexVal);
printf("Tada: %d\n", indexVal);
return 0;
}