HDOJ 1212 Big Number（大数同余）

最新推荐文章于 2020-05-04 19:26:48 发布

CHRISTINA_ERZA

最新推荐文章于 2020-05-04 19:26:48 发布

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分类专栏：大数

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本文介绍了一种处理大数取模运算的算法实现，并通过具体示例演示了如何使用两个同余定理公式来简化计算过程。算法适用于ACM竞赛等场景。

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Big Number

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

Total Submission(s): 7269 Accepted Submission(s): 5022

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

这一题用了以下两个同余定理公式：

(A + B) mod M = ( A mod M + B mod M ) mod M

(A * B) mod M = ((A mod M) *( B mod M)) mod M

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