HDU 5832：A water problem （大数整除）

发布时间：2021-01-24 07:43:49 所属栏目：大数据来源：网络整理

导读：A water problem Time Limit: 5000/2500 MS (Java/Others)????Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 412????Accepted Submission(s): 218 Problem Description Two planets named Haha and Xixi in the universe and they were

A water problem

Time Limit: 5000/2500 MS (Java/Others)????Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 412????Accepted Submission(s): 218

Problem Description Two planets named Haha and Xixi in the universe and they were created with the universe beginning.

There is? 73 ?days in Xixi a year and? 137 ?days in Haha a year.?

Now you know the days? N ?after Big Bang,you need to answer whether it is the first day in a year about the two planets. ?
Input There are several test cases(about? 5 ?huge test cases).

For each test,we have a line with an only integer? N(0≤N) ,the length of? N ?is up to? 10000000 . ?
Output For the i-th test case,output Case #i:,then output "YES" or "NO" for the answer. ?
Sample Input

?
Sample Output

  
  
   
   Case #1: YES
Case #2: YES
Case #3: NO

?
Author UESTC ?

题意：判断输入的数是否可以同时整除73与137，也就是能不能整除73与137的最小公倍数。
因为输入的数的长度是10000000以内，我们只能采用字符串来存储它，从ans初始化为0，最低位开始ans=(ans+s[i]-'0')%b;?计算到最后，如果ans?等于0，可以整除，否则，不可以整除。

AC代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>

using namespace std;

char s[11000000];
int ans,b=73*137;
int main()
{
    for(int i=1; gets(s); i++)
    {
        ans=0;
        int l=strlen(s);
        for(int j=0; j<l; j++)
            ans=(ans*10+(s[j]-'0'))%b;
        if(!ans)printf("Case #%d: YESn",i);
        else printf("Case #%d: NOn",i);
    }
    return 0;
}

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