A water 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.
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.
10001 0 333
Sample Output
Case #1: YES Case #2: YES Case #3: NO
题目连接:
http://acm.hdu.edu.cn/showproblem.php?pid=5832
题意
给你一个数,问你这个数是否能够整除137和73
题解:
这两个数互质,其实就是问你能否整除10001(73*137,)
这个直接扫一遍就好了。
用string可能会TLE,所以老老实实用char就好了
Java 都是 MLE,看来Java也不是万能的,
AC代码:
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
char a[10000005];
int main()
{
int kase=0;
int n;
//freopen("data/5832.txt","r",stdin);
while(cin>>a)
{
printf("Case #%d: ",++kase);
int len=strlen(a);
int mod=0;
for(int i=0;i<len;i++)
{
mod=(mod*10+a[i]-'0')%10001;
}
if(mod==0)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}
Java MLE 代码:
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
int kase = 0;
while(sc.hasNext()){
BigInteger a = sc.nextBigInteger();
BigInteger mod = BigInteger.valueOf(10001);
BigInteger ans = a.mod(mod);
System.out.printf("Case #%d: ",++kase);
if(ans.equals(BigInteger.ZERO)){
System.out.println("YES");
}else{
System.out.println("NO");
}
}
}
}
(编辑:晋中站长网)
【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容!
|