Big Number
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.
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
代码：
1 #include<cstdio> 2 #include<cstring> 3 4 int main() 5 { 6 int a[1005]; 7 char s[1005]; 8 int m,i,j; 9 while(scanf("%s%d",s,&m)!=EOF) 10 { 11 int len=strlen(s); 12 for(j=0,i=len-1;i>=0;i--) 13 a[j++]=s[i]-'0'; 14 int sum=0; 15 for(j-=1;j>=0;j--) 16 sum=(sum*10+a[j])%m; 17 printf("%d ",sum); 18 } 19 return 0; 20 }