HDU 5832 A water problem 大数取余

A water problem

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 650    Accepted Submission(s): 333

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

 Source

2016中国大学生程序设计竞赛 - 网络选拔赛

 

 

ccpc的网络赛被虐的好惨啊 做完水题其他就下不了手了

还要加油  

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN     freopen("input.txt","r",stdin);
#define FOUT    freopen("output.txt","w",stdout);
#define INF     0x3f3f3f3f
#define INFLL   0x3f3f3f3f3f3f3f
#define lson    l,m,rt<<1
#define rson    m+1,r,rt<<1|1
typedef long long LL;
typedef pair<int,int> PII;

const int MAXN = 10000005;

char s[MAXN];

int check(char s[], int mod){
    
    //将数字看成多项式的和  比如12345就是1*10^5+2*10^4+3*10^3+4*10^2+5*10^1
    //先把数求模和分开求再合起来求模结果是一样的
    
    LL sum = 0;
    int len = strlen(s);
    for( int i = 0; i < len ; i++){
        sum = sum*10 + s[i]-48;
        sum %= mod;
    }
    return sum;
}

int main()
{
    //FIN
    int cas = 1;
    while(~scanf("%s", s)){
        if(check(s, 137) == 0 && check(s, 73) == 0)  printf("Case #%d: YES
",cas++);
        else  printf("Case #%d: NO
",cas++);
    }

}