HDU 1212 Big Number(数论)

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2407    Accepted Submission(s): 1639

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

Author

Ignatius.L

Source

杭电ACM省赛集训队选拔赛之热身赛

 

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Eddy

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解题报告：这是一道数论题！就是求AmodB，但是A是大数，想了好半天，也没有想出神马思路，就借鉴了一下网上“大牛”们的思路

下面就是我在网上找的简单方法：

用递推的思想，就可以总结出下面的公式；

举例：

12345 9

余数等于（12340%9+5%9）%9；

而12340 9

（12300%9+40%9）%9;

依次...

最后（10000%9+2000%9）%9；

而10000%9=（1%9*10000）%9

2000%9=（2%9*1000）%9

即（1*10+2）%9*1000%9；

即可得到：for（i=0;i<len;i++）

{

sum=sum*10+s[i]-'0';

sum=sum%9;

}

我的代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 1010;
char str[N];
int main()
{
    int ans,i,len,m;
    while(scanf("%s%d",str,&m)!=EOF)
    {
        len=strlen(str);
        ans=0;
        for(i=0;i<len;i++)
        {
            ans=ans*10+str[i]-'0';
            ans=ans%m;

        }
        printf("%d\n",ans);
    }
    return 0;
}