Problem Description
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 3 12 7 152455856554521 3250
Sample Output
2 5 1521
View Code
1 #include<stdio.h> 2 #include<string.h> 3 char A[10005]; 4 int n; 5 int main() { 6 while(scanf("%s%d",A,&n)!=EOF) { 7 int len=strlen(A); 8 int sum=0; 9 for(int i=0;i<len;i++) { 10 sum=10*sum+A[i]-'0'; 11 sum=sum%n; 12 } 13 printf("%d\n",sum); 14 } 15 return 0; 16 }