Hdu2053

Switch Game HDU - 2053

Time limit1000 msMemory limit32768 kB

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 ).

InputEach test case contains only a number n ( 0< n<= 10^5) in a line.
OutputOutput the condition of the n-th lamp after infinity operations ( 0 - off, 1 - on ).Sample Input

1
5

Sample Output

1
0


首先我们得知道，第n次开关改变n的倍数的灯，那么前n盏灯的状态不再改变。所以要知道无限次操作后第n盏灯的状态，只需计算n盏灯n次操作后的状态。
一开始想到的是模拟。时限是一秒，1秒大概能做10^8次操作，而题目的范围是10^10，不出意外TLE了。

#include <cstdio>
#include <iostream>
#include <queue>
#include <vector>
#include<string.h>
#include<map>
#include<bits/stdc++.h>
#define LL long long
#define maxn 100005
using namespace std;
int a[maxn];
int n;
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
     memset(a,0,sizeof a);
     for(int i=1;i<=n;i++)
     {
         for(int j=1;j<=n;j++)
         {
             if((j%i)==0)
                a[j]=!a[j];
         }
     }
     printf("%d\n",a[n]);
    }
    return 0;
}

接着，想到第n盏的开关实际决定与它因数的个数，偶数个因数-0，奇数个因数-1.

#include <cstdio>
#include <iostream>
#include <queue>
#include <vector>
#include<string.h>
#include<map>
#include<bits/stdc++.h>
#define LL long long
#define maxn 100005
using namespace std;
int n,ans;
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        ans=0;//记得初始化
        for(int i=1;i<=n;i++)
        if((n%i)==0)
            ans++;
        if(ans&1)printf("1\n");
        else printf("0\n");
    }
    return 0;
}