hdu 2053 Switch Game 水题一枚，鉴定完毕

Switch Game

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10200    Accepted Submission(s): 6175

Problem Description

There are many lamps in a line. All of them are off at first. A series of operations are carried out on these lamps. On the i-th operation, the lamps whose numbers are the multiple of i change the condition ( on to off and off to on ).

 

 

Input

Each test case contains only a number n ( 0< n<= 10^5) in a line.

 

 

Output

Output the condition of the n-th lamp after infinity operations ( 0 - off, 1 - on ).

 

 

Sample Input

1 5

 

 

Sample Output

1 0

Hint

hint

Consider the second test case:

       The initial condition : 0 0 0 0 0 …

    After the first operation : 1 1 1 1 1 …

After the second operation : 1 0 1 0 1 …

   After the third operation : 1 0 0 0 1 …

 After the fourth operation : 1 0 0 1 1 …

   After the fifth operation : 1 0 0 1 0 …

The later operations cannot change the condition of the fifth lamp any more. So the answer is 0.

 

 

Author

LL

 

 

转化求n个因子个数，暴力0ms过了(⊙o⊙)…

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;

int num_Euler(int n)
{
    int i;
    int num,ans=1;
    for(i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            num=1;
            while(n%i==0)
            {
                n=n/i;
                num++;
            }
            ans=ans*num;
        }
    }
    if(n!=1)
        ans=ans*2;
    return ans;
}
int main()
{
    int n;
    while(scanf("%d",&n)>0){
        int m=num_Euler(n);
        printf("%d\n",(m&1));
    }
    return 0;
}

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;

bool dp[100002];

void init()
{
    int i,j;
    memset(dp,true,sizeof(dp));
    for(i=2;i<=100000;i++)
    {
        for(j=i;j<=100000;j=j+i)
            if(dp[j]==true) dp[j]=false;
            else dp[j]=true;
    }
}
int main()
{
    int n;
    init();
    while(scanf("%d",&n)>0)
    {
        if(dp[n]==false)printf("0\n");
        else printf("1\n");
    }
    return 0;
}