hdu 5795 A Simple Nim (sg函数)

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 1≤T≤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≤n≤106,1≤s[i]≤109)

 

Output

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

 

Sample Input

2

2

4 4

3

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
", 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.");
    }
}