Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5930 Accepted Submission(s): 4146
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
此题需要用到同余定理：
常用的同余定理有以下三种：
（1）（（A mod m）*（B mod m））mod m ==（A*B）mod m
（2）（（A mod m）+（B mod m））mod m ==（A+B）mod m
（3）（（A mod m）-（B mod m）+ m）mod m ==(A-B)mod m
在减法中，由于a mod n 可能小于b mod n，需要在结果上加上n.
我们先了解下如何运用同余定理求余数：
对大数求余数：
/*例如10000对m求余：
* 10000%m
* ==(10%m*1000%m)%m
* ==(10%m*(10%m*100%m)%m)%m
* ==(10%m*(10%m*(10%m*10%m)%m)%m)%m
* 用代码表示就是:
* 假设10000是字符串长度是len
*/
/*
* 如123对m求余
* 123%m
* ==((12%m*10%m)%m+3%m)%m
* ==(((10%m+2%m)%m*10%m)%m+3%m)%m
* ==((((1%m*10%m)%m+2%m)%m*10%m)%m+3%m)%m
*/
gets(str);
int ans=0;
for(i=0;i<len;i++)
{
ans=ans*10+str[i];
ans=ans%m;
}
对幂求余数：
/*
* 对幂取模如对37的4次方取模
* （37*37*37*37）%m
* ==(37%m*(37*37*37)%m)%m
* ==(37%m*(37%m*(37*37)%m)%m)%m
* ==(37%m*(37%m*(37%m*37%m)%m)%m)%m
*/
//求n^m%1000
s=n;
for(i=1;i<m;i++)
{
s=s*n;
s=s%1000;
}
此题用到了（1）（2）两个：
#include<stdio.h>
#include<string.h>
#define MAX 1100
int main()
{
int n,m,j,i,s,t;
char p[MAX];
while(scanf("%s",p)!=EOF)
{
scanf("%d",&n);
int l=strlen(p);
s=0;
for(i=0;i<l;i++)
{
s=s*10+p[i]-'0';
s=s%n;
}
printf("%d\n",s);
}
return 0;
}
