HDU 1212 Big Number （大数取模）

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

2 312 7152455856554521 3250

Sample Output

251521

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 50000char 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;}