hdu 5742 chess SG函數

Chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1908    Accepted Submission(s): 831

Problem Description

Alice and Bob are playing a special chess game on an n × 20 chessboard. There are several chesses on the chessboard. They can move one chess in one turn. If there are no other chesses on the right adjacent block of the moved chess, move the chess to its right adjacent block. Otherwise, skip over these chesses and move to the right adjacent block of them. Two chesses can’t be placed at one block and no chess can be placed out of the chessboard. When someone can’t move any chess during his/her turn, he/she will lose the game. Alice always take the first turn. Both Alice and Bob will play the game with the best strategy. Alice wants to know if she can win the game.

 

Input

Multiple test cases.

The first line contains an integer T(T≤100), indicates the number of test cases.

For each test case, the first line contains a single integer n(n≤1000), the number of lines of chessboard.

Then n lines, the first integer of ith line is m(m≤20), indicates the number of chesses on the ith line of the chessboard. Then m integers pj(1≤pj≤20)followed, the position of each chess.

 

Output

For each test case, output one line of “YES” if Alice can win the game, “NO” otherwise.

 

Sample Input

2 1 2 19 20 2 1 19 1 18

 

Sample Output

NO YES

題意：給n堆石子，有兩種操作，第一種是選擇任意一堆取走一些石子，可以取完這一堆，但不能不取。第二種操作是把某一堆分成三堆石子，每堆石子不能爲空。問先手是否必勝。

SG函數，因爲數據是1e9，所以打表找一下規律。

#include <bits/stdc++.h>

using namespace std;

int SG[1000];
int sg(int x)
{
    if(SG[x] != -1) return SG[x];
    int vis[150] = {0};
    for(int i = 1;i <= x;i++)     ///正常的nim取石子
        vis[sg(x-i)] = 1;
    for(int i = 1;i <= 100;i++){  ///採用另一種操作，分成三堆
        for(int j = 1;j <= 100;j++){
            for(int k = 1;k <= 100;k++){
                if(i+j+k == x){
                    vis[sg(i)^sg(j)^sg(k)] = 1;
                }
            }
        }
    }
    for(int i = 0;;i++)
        if(!vis[i]) return SG[x] = i;
}

void Init()
{
    memset(SG,-1,sizeof SG);
    SG[0] = 0;
    SG[1] = 1;
    SG[2] = 2;
    sg(100);
    for(int i = 0;i <= 100;i++) printf("%d %d\n",i,SG[i]);
}

int SG_(int x)
{
    if(x%8 == 7) return x+1;
    if(x%8 == 0) return x-1;
    return x;
}

int main()
{
    //Init();  打表找規律
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        scanf("%d",&n);
        int ans = 0;
        for(int i = 1;i <= n;i++){
            int t;
            scanf("%d",&t);
            ans ^= SG_(t);
        }
        if(ans) puts("First player wins.");
        else    puts("Second player wins.");
    }
    return 0;
}