Big Number

最新推荐文章于 2021-08-19 13:51:23 发布

原创最新推荐文章于 2021-08-19 13:51:23 发布 · 177 阅读

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本文介绍了一种处理大数取余问题的高效算法，通过逐位处理和利用同余定理，实现对大数A与较小数B进行取余运算。示例代码展示了如何读取输入的大数和较小数，并通过算法计算出结果。

Big Number

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

大概题意是一个大数与第二个数取余，同余定理(a%c + b%c)%c = (a + b) %c即可

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

int main()
{
    char a[1005];
    int b;
    while(scanf("%s %d", &a, &b) != EOF)
    {
        int i, ans = 0;
        for(i = 0; a[i] != '\0'; i++)
        {
            ans = ((ans * 10) % b + (a[i] - '0') % b) % b;
        }
        cout << ans <<endl;
    }
    return 0;
}