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    There are two seagulls playing a very peculiar game. First they line up N unit squares in a line, all originally colored white.

    Let L be the length of the longest continuous sub-segment of white unit squares. Let P be any prime that satisfies the condition that . The player can choose any P if it satisfies the conditions but if there is no value of P that does, then P is set to 1.

    The current player then proceeds to choose any continuous sub-segment of white unit squares of length P and paint them black. The player who can’t make a move loses.

    If both players play optimally and the first player starts, which player is going to win the game?

    Input

    The first line of input is T – the number of test cases.

    Each test case contains a line with a single integer N (1 ≤ N ≤ 107).

    Output

    For each test case, output on a single line "first" (without quotes) if the first player would win the game, or "second" if the second would win.

    Example

    Input
    2
    2
    5
    Output
    second
    first

    题意:
    每次可以取一段白色连续区间内p个,把白的染成黑色。不能染色的玩家失败。
    整段区间内,最长的白色连续区间的长度为L,p是小于等于L/2的任意质数,如果没有质数,p可以取1;
    思路:
    手推,2,3是先手必败,1是先手必胜。
    大于3的偶数,在中间取2,使留下来的两段长度相等。
    大于3的奇数,在中间取3,使留下来的两段长度相等。
    然后就像nim博弈只有两堆石子那样,跟着第二个人取,就可以坐等胜利了。
    #include<cstdio>
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--){
            int n;
            scanf("%d",&n);
            if(n==2||n==3){printf("second
    ");}
            else printf("first
    ");
        }
        return 0;
    }
    View Code
    
    
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  • 原文地址:https://www.cnblogs.com/ZGQblogs/p/10661452.html
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