hdoj 1212 Big Number 【同余定理】

Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6776 Accepted Submission(s): 4678



[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

题意：给出a，b，求a对b取余的结果，a是大数

代码：

#include<stdio.h>
#include<string.h>
char a[20000000];
int main()
{
int i,l;
int b;
while(scanf("%s%d",a,&b)!=EOF)
{
long long sum=0;
l=strlen(a);
for(i=0;i<l;i++)
{
sum=(sum*10+a[i]-'0')%b;
}
printf("%lld\n",sum);
}
return 0;
}