HDU 1212 Big Number（大数取模）

最新推荐文章于 2021-08-19 13:51:23 发布

Spectacules

最新推荐文章于 2021-08-19 13:51:23 发布

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分类专栏： ACM__数论&&数学技巧

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本文介绍了一种解决大数取模问题的高效算法，通过按位取模的方法简化了计算过程，适用于需要处理超大整数场景的ACM编程。

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Big Number

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

【思路分析】
用了大数取模最基本的方法——按位取模。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 100005;
int mod,len,ans;
char s[maxn];
void init()
{
    len = strlen(s);
    ans = 0;
}
void solve()
{
    for(int i = 0;i < len;i++)
    {
        ans = (ans * 10 + (s[i] - '0') % mod) % mod;
    }
    printf("%d\n",ans);
}
int main()
{
    while(scanf("%s%d",s,&mod) != EOF)
    {
        init();
        solve();
    }
    return 0;
}