HDU_1212 Big Number

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本文介绍了一种处理大数取模问题的有效算法，并通过实例详细解释了如何利用同余定理进行快速计算。该算法特别适用于ACM竞赛等场景。

Big Number

2014-8-4 15:46

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

同余定理

/*
举个例子，1314 % 7= 5
由秦九韶公式：1314= （（1*10+3）*10+1）*10+4
所以有 1314 % 7= （（（1 * 10 % 7 +3 ）*10 % 7 +1）*10 % 7 +4 ）%7
*/

#include<stdio.h>
#define maxn 1010
char str[maxn];
int main()
{
    int n,i,ans;
    while(~scanf("%s%d",str,&n))
    {
        ans=0;
        for(i=0;str[i]!='\0';i++)
        {
            ans=(ans*10+(str[i]-'0')%n)%n;
        }
        printf("%d\n",ans);
    }
    return 0;
}