hdoj1212Big Number

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case, you have to ouput the result of A mod B.

[align=left]Sample Input[/align]

2 3
12 7
152455856554521 3250

[align=left]Sample Output[/align]

2
5
1521代码：#include<stdio.h>
int main()
{
char a[1005];
int b,c;
while(scanf("%s %d",a,&b)!=EOF)
{
long long sum=0;
for(int i=0;a[i]!='\0';i++)
{
sum=(sum*10+a[i]-'0')%b;
}
printf("%d\n",sum%b);
}
return 0;
}
思路：每一步都去余，防止爆掉