hdu 5832 A water problem【大数取膜】

大数取模判断题解法

最新推荐文章于 2018-08-26 19:38:55 发布

原创最新推荐文章于 2018-08-26 19:38:55 发布 · 677 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#hdu 5832 #杭电 5832

水题专栏收录该内容

175 篇文章

订阅专栏

A water problem

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 166 Accepted Submission(s): 87

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

10001

333

Sample Output

Case #1: YES

Case #2: YES

Case #3: NO

Author

UESTC

题目大意：

给你一个数，问你这个数能否整除73和137.

思路：

注意N的范围比较大，需要使用大数取膜的做法。

我们定义一个sum=0，每一位sum=sum*10+当前位数。然后将sum%=73.当然137同理。如果最终得到的sum==0，即是说明这个数可以整除73。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define ll __int64
char a[10000030];
int main()
{
    int kase=0;
    while(~scanf("%s",a))
    {
        ll sum=0;
        ll sum2=0;
        int n=strlen(a);
        for(int i=0;i<n;i++)
        {
            sum=(sum*10+a[i]-'0')%73;
            sum2=(sum2*10+a[i]-'0')%137;
        }
        printf("Case #%d: ",++kase);
        if(sum%73==0&&sum%137==0)
        {
            printf("YES\n");
        }
        else printf("NO\n");
    }
}