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.

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.

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

2 3
12 7
152455856554521 3250

2
5
1521

Ignatius.L

Eddy | We have carefully selected several similar problems for you: 1215 1211 1210 1214 1695

#include<stdio.h> 
#include<string.h>
int main(){
	int b;
	char a[1010];
	while(~scanf("%s%d",a,&b)){
		 int len=strlen(a),res=0;
		 for(int i=0;i<len;++i){
		 	res=res*10+a[i]-'0';
		 	if(res>=b) 
		 	res=res%b;
		 }
	     printf("%d\n",res);
	}
	return 0;
}