hdu 2053 Switch Game

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Switch Game Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8850    Accepted Submission(s): 5286 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 Hinthint 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.   题意:就是初始时有很多灯的状态都是关闭的, 然后第n次改变处于n的倍数的位置的灯的状态, 即:由0到1,由1到0; 思路:以下是我的代码,直接循环,当j大于n时, 即再也无法改变其状态,所以可以循环到n结束。 但是还可以找规律:只有处于n*n状态的灯才会是亮的。 #include <iostream> #include <cstring> #include <cstdio> #include<cmath> #include <algorithm> using namespace std; int main() { int n,i,j,a[100010]; while(~scanf("%d",&n)) { memset(a,0,sizeof(a)); int j=1; while(j<=n) { for(i=j;i<=n+2;i+=j) a[i]=!a[i]; j++; } //for(i=1;i<=n;i++) //cout<<a[i]<<' '; printf("%d\n",a[n]); } return 0; }

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