Big Number（1212）

As we know, Big Number is always troublesome. But it's really important in our ACM. And today, your task is to write a program to calculate A mod B.

To make the problem easier, I promise that B will be smaller than 100000.

Is it too hard? No, I work it out in 10 minutes, and my program contains less than 25 lines.

 

Input

The input contains several test cases. Each test case consists of two positive integers A and B. The length of A will not exceed 1000, and B will be smaller than 100000. Process to the end of file.

 

Output

For each test case, you have to ouput the result of A mod B.

 

Sample Input

2 312 7152455856554521 3250

 

Sample Output

25
1521
秦九韶
#include <stdio.h> #include <string.h>  #include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <cctype>#include <iostream>#include <algorithm>#include <vector>#include <stack>#include <deque>#include <queue>#include <set>#include <list> #include <map> #include <string>using namespace std; #define INF 2147483647int main(){int n,i,l,sum;char s[100000];while(scanf("%s %d",s,&n)!=EOF){l=strlen(s);sum=0;for(i=0;i<l;i++){sum*=10;sum+=s[i]-'0';sum%=n;}cout <<sum <<endl;}return 0;}