hdu 5088 高斯消元n堆石子取k堆石子使剩余异或值为0

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http://acm.hdu.edu.cn/showproblem.php?pid=5088

求能否去掉几堆石子使得nim游戏胜利

我们可以把题目转化成求n堆石子中的k堆石子数异或为0的情况数。使用x1---xn表示最终第i堆石子到底取不取（1取，0不取），将每堆石子数画成2进制的形式，列成31个方程来求自由变元数，最后由于自由变元能取1、0两种状态，所以自由变元数多于0即可输出Yes。

注意有40+个方程，因为A[I]<=1e12....

#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <string>#include <queue>#include <map>#include <iostream>#include <algorithm>using namespace std;#define RD(x) scanf("%d",&x)#define RD2(x,y) scanf("%d%d",&x,&y)#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define clr0(x) memset(x,0,sizeof(x))#define clr1(x) memset(x,-1,sizeof(x))#define eps 1e-9const double pi = acos(-1.0);typedef long long LL;typedef unsigned long long ULL;const int modo = 1e9 + 7;const int INF = 0x3f3f3f3f;const int inf = 0x3fffffff;const LL _inf = 1e18;const int maxn = 1005,maxm = 10005;#define MAXN 1100#define MOD 1000007#define weishu 42LL a[MAXN], g[MAXN][MAXN];int Gauss(int n) {    int i, j, r, c, cnt;    for (c = cnt = 0; c < n; c++) {        for (r = cnt; r < weishu; r++) {            if (g[r][c])                break;        }        if (r < weishu) {            if (r != cnt) {                for (i = 0; i < n; i++)                    swap(g[r][i], g[cnt][i]);            }            for (i = cnt + 1; i < weishu; i++) {                if (g[i][c]) {                    for (j = 0; j < n; j++)                        g[i][j] ^= g[cnt][j];                }            }            cnt++;        }    }    return n - cnt;}int main() {    int c;    int n, i, j;    int ans, vary;    scanf("%d", &c);    while (c--) {        int fuck = 0;        scanf("%d", &n);        for (i = 0; i < n; i++){            scanf("%I64d", &a[i]);            fuck ^= a[i];        }        for (i = 0; i < weishu; i++) {            for (j = 0; j < n; j++)                g[i][j] = (a[j] >> i) & 1;        }        vary = Gauss(n);        //printf("%d\n", vary);        if(vary <= 0 )//|| (fuck == 0 && vary <= 1))            puts("No");        else            puts("Yes");    }    return 0;}

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