（HDU）5832 - A water problem【大数取模】

A water problem

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1851 Accepted Submission(s): 739

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
0
333

Sample Output

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

Author

UESTC

题意：判断一个大数能否同时被73和137整除。

分析：大数取模直接水过去，把大数拆成一位位的取模相加。

#include <bits/stdc++.h>

using namespace std;

int main()
{
    char s[10000010];
    int i,len,ans,kase=0;
    while(~scanf("%s",s))
    {
        len=strlen(s),ans=0;
        for(i=0;i<len;i++)
        {
            ans=10*ans+s[i]-'0';
            ans%=10001;
        }
        printf("Case #%d: ",++kase);
        if(ans) printf("NO\n");
        else printf("YES\n");
    }
    return 0;
}

这题因为73和137互质，所以只要对他们乘积10001取模就可以了，正常套路是下面的代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")  
#include<stdio.h>  
#include<string.h>  
#include<stdlib.h>  
#include<queue>  
#include<stack>  
#include<math.h>  
#include<vector>  
#include<map>  
#include<set>  
#include<cmath>  
#include<complex>  
#include<string>  
#include<algorithm>  
#include<iostream>  
#define eps 1e-9  
#define LL long long  
#define bitnum(a) __builtin_popcount(a)  
using namespace std;  
const int N = 10000005;  
const int M = 10005;  
const int inf = 1000000007;  
const int mod = 1000000007;  
char s[N];  
int main()  
{  
    int ans1,ans2,p=1,i,k;  
    while(~scanf("%s",s))  
    {  
        ans1=ans2=0;  
        k=strlen(s);  
        for(i=0;i<k;i++)  
        {  
            ans1=(ans1*10+s[i]-'0')%73;  
            ans2=(ans2*10+s[i]-'0')%137;  
        }  
        if(!ans1&&!ans2)  
            printf("Case #%d: YES\n",p++);  
        else  
            printf("Case #%d: NO\n",p++);  
    }  
    return 0;  
}