HDU2053Switch Game

Switch Game

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13114    Accepted Submission(s): 7971

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

15

 

Sample Output

10
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

 

Source

校庆杯Warm Up

 

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            简单题：题目乍一看好像很难，但再仔细看看题不难发现其实第n个灯泡在第n次操作后的状态都是基于其之前的因子数个数，那问题到这边很简单了。

求因子数个数如果实在不会，由于本题的数据并不大，大可暴力检索。这里我用到的是将1000000范围内整数因子数统一打表，在不包含自身的条件下:

如果因子数个数为奇数输出 0 即可！如果不是就输出 1

#include<iostream>#include<cstring>#define N 1000000using namespace std;int dp[N];void gcd(){    memset(dp,0,sizeof(dp));    int i,j;    int m=N/2;    for(i=1;i<m;i++)    for(j=i+i;j<N;j=j+i)      dp[j]++;}int main(){gcd();int n;    while(cin>>n){    if(dp[n]%2) cout<<"0"<<endl;    else cout<<"1"<<endl;}return 0;}