HDU 1212 Big Number （大数取模）

最新推荐文章于 2021-04-18 09:48:20 发布

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最新推荐文章于 2021-04-18 09:48:20 发布

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分类专栏： HDUOJ ACM_高精度

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本文介绍了一种解决大数求模问题的有效算法，并提供了一个简洁的C语言实现示例。通过对输入的大数A和较小的正整数B进行逐位处理，避免了使用长整型数据类型可能引发的问题。

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Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7103 Accepted Submission(s): 4898

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

Author

Ignatius.L

Source

杭电ACM省赛集训队选拔赛之热身赛

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题意：大数求模。

题解：看代码吧。。。

AC代码：

#include <stdio.h>
#include<string.h>
#include <stdlib.h>
#define N 50000
char s[N];
int main()
{
    int mod,m;
    while(~scanf("%s %d",s,&m))
   {
       mod=0;
       for(int i=0;i<strlen(s);i++)
       mod=(mod*10+s[i]-'0')%m;
       printf("%d\n",mod);
   }
    return 0;
}

WA，爆LL

WA代码：

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long LL;

int mod(LL a,  LL b)
{
	int ans=a;
	while(a>b)
	{
		a=a%b;
		ans=a;
	}
	return ans;
}

int main()
{
    ios::sync_with_stdio(false);
    LL n,m;
    int res;
    while(cin>>n>>m)
    {
    	res=mod(n,m);
    	printf("%d\n",res);
	}
	return 0;
}