HDU2176 取(m堆)石子游戏

来源：互联网发布：python 判断加载完成编辑：程序博客网时间：2024/05/29 03:28

链接：http://acm.hdu.edu.cn/showproblem.php?pid=2176

题意：nim游戏，先手必败输出no，先手必胜输出yes和第一步的每一种取法。

思路：基础nim，顺序输出取法就行。

这是篇nim博弈入门的经典帖子http://acm.hdu.edu.cn/forum/read.php?fid=9&tid=10617

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<cmath>#include<set>using namespace std;int a[200005];int main(){    int m;    while (scanf("%d", &m), m){        int c = 0;        for (int i = 0; i < m; i++){            scanf("%d", &a[i]);            c ^= a[i];        }        if (c == 0){            printf("No\n");            continue;        }        printf("Yes\n");        for (int i = 0; i < m; i++){            if ((a[i] ^ c) < a[i])                printf("%d %d\n", a[i], a[i] ^ c);        }    }    return 0;}

因为nim博弈的性质不熟悉超时一发，不需要每个数都判断是否含有c的最高位，只要含有c的最高位，与c亦或一定小于原数。

超时代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<cmath>#include<set>using namespace std;int a[200005];int main(){    int m;    while (scanf("%d", &m), m){        int c = 0;        for (int i = 0; i < m; i++){            scanf("%d", &a[i]);            c ^= a[i];        }        if (c == 0){            printf("No\n");            continue;        }        printf("Yes\n");        int cc = 1;        while ((cc << 1) <= c)cc <<= 1;        int t = cc << 1;        for (int i = 0; i < m; i++){            if (a[i] % t >= cc)                printf("%d %d\n", a[i], a[i] ^ c);        }    }    return 0;}

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