HDU 1212 Big Number（简单的大数取模）

发布时间：2020-12-14 02:24:33 所属栏目：大数据来源：网络整理

导读：链接：http://acm.hdu.edu.cn/showproblem.php?pid=1212 Big Number Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6037????Accepted Submission(s): 4226 Problem Description As we know,Bi

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1212

Big Number

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6037????Accepted Submission(s): 4226

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

题意：求长度不大于1000的a模于b(b<=100000)。

解题思路：题目看似很难，其实这是一道很简单的大数求模。首先要先有数学知识：a,b同余有，a%m = b%m ；? (a±b)%m = (a%m±b%m)%m ；(a*b)%m = (a%m*b%m)%m 。所以：将大数转化为字符串处理，然后问题可以等价于各个位数上的分别取模。

<span style="font-size:18px;">#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn=1010;
char a[maxn];
int m,b[maxn];
int gcd(char *a,int m)
{
    int len=strlen(a);
    for(int i=0;i<len;i++) b[i]=a[i]-'0';
    int ans=0;
    for(int i=0;i<len;i++)
    {
        ans=(ans*10+b[i])%m;
    }
    return ans;
}
int main()
{
    while(cin>>a>>m)
    {
        cout<<gcd(a,m)<<endl;
    }
    return 0;
}
</span>

（编辑：李大同）

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