hdu 5795 A Simple Nim (sg函数)-优快云博客

本文探讨了ASimpleNim游戏的策略问题，这是一种两人轮流从多堆糖果中取糖果的游戏，可以拆分堆但不能不取。文章通过分析找到了获胜的关键规律，并给出了实现代码。

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A Simple Nim

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 33 Accepted Submission(s): 18

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

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.

找到规律以后很好做，关键是打表的程序。

打表程序

#include <bits/stdc++.h>
using namespace std;
int g[1010];
void init() {
    memset(g, -1, sizeof(g));
}
int getSG(int x) {
    if(g[x] != -1) return g[x];
    if(x == 0) return 0;
    if(x == 1) return 1;
    if(x == 2) return 2;
    int vis[110];
    memset(vis, 0, sizeof(vis));
    for(int i = 1; i < x; i++) {
        int t = 0;
        int a = getSG(i);
        t ^= a;
        for(int j = 1; j < x - i; j++) {
            int tt = t;
            int b = getSG(j);
            int c = getSG(x - i - j);
            tt ^= b;
            tt ^= c;
            vis[tt] = 1;
            vis[c] = vis[b] = 1;
        }
        vis[a] = 1;
        vis[t] = 1;
    }
    vis[0] = 1;
    for(int i = 0; ; i++) if(!vis[i])
        return i;
}
int main() {
    int n;
    init();
    for(int i = 1; i <= 100; i++) {
        g[i] = getSG(i);
        printf("%d %d %d\n", i, i % 8, g[i]);
    }
}

ac代码

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
int getSG(int x) {
    if(x == 0)
        return 0;
    if(x % 8 == 0)
        return x - 1;
    if(x % 8 == 7)
        return x + 1;
    return x;
}

int a[maxn];

int main() {
    int t;
    scanf("%d", &t);
    while(t--) {
        int n;
        scanf("%d",&n);
        int ans = 0;
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            ans ^= getSG(a[i]);
        }
        if(!ans)
            puts("Second player wins.");
        else
            puts("First player wins.");
    }
}

转载于:https://www.cnblogs.com/lonewanderer/p/5737674.html