A Simple Nim

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 980 Accepted Submission(s): 573

Problem Description

Two players take turns picking candies from n heaps,the player who picks the last one will win the game.On each turn they can pick any number of candies which come from the same heap(picking no candy is not allowed).To make the game more interesting,players can separate one heap into three smaller heaps(no empty heaps)instead of the picking operation.Please find out which player will win the game if each of them never make mistakes.

Input

Intput contains multiple test cases. The first line is an integer $1\le T\le 100$, the number of test cases. Each case begins with an integer n, indicating the number of the heaps, the next line contains N integers $s[0],s[1], ....,s[n-1]$, representing heaps with $s[0],s[1], ...,s[n-1]$ objects respectively.$(1\le n\le 10^6,1\le s[i]\le 10^9)$

Output

For each test case,output a line whick contains either"First player wins."or"Second player wins".

Sample Input

4 4

1 2 4

Sample Output

Second player wins.

First player wins.

题意：n堆石子，两个人轮流拿，每次可以选择任意一堆取任意个（不能不拿）或者将一个堆分成3个小堆，问先手胜还是后手胜。

题解：SG定理的应用，可以看做是Nim博弈，打表找规律，我也不会证

　　i=8*k+7时SG[i]=8*k+8，i=8*k+8时SG[i]=8*k+7其他情况SG[i]=i。

#include<bits/stdc++.h>

#define N 45000

#define mes(x) memset(x, 0, sizeof(x));

#define ll __int64

const long long mod = 1e9+;

const int MAX = 0x7ffffff;

using namespace std;

int dir[], SG[];

int getSG(int n){

    if(n == ) return ;

    if(n% == ) return n-;

    if(n% == ) return n+;//注意这里是n%8==7,不是n%7==0....

    return n;

}

int main()

{

    int T,t, i,n, ans;

    scanf("%d", &T);

    while(T--){

        scanf("%d", &n);

        for(ans=i=;i<n;i++){

            scanf("%d", &t);

            ans ^= getSG(t);

        }

        if(ans==) printf("Second player wins.\n");

        else printf("First player wins.\n");

    }

    return ;

}

巴特西

HDU5795A Simple Nim SG定理

A Simple Nim

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