HDU 3915 Game (高斯消元)
题意：有n堆石子，每个人只能从某一堆至少拿走一个，不能拿者败。问事先拿走某些堆的石子，使得先手必败。
析：将石子拆成二进制，未知数为1表示保留该堆石子，为0表示事先拿走该堆石子。最后求自由变元的数目，就是2的幂。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 10;
const int mod = 1000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int equ, var, free_num;
bool a[maxn][maxn];
int Gauss(){
int max_r, col, k;
free_num = 0;
for(k = col = 0; k < equ && col < var; ++k, ++col){
max_r = k;
for(int i = k+1; i < equ; ++i)
if(a[i][col] > a[max_r][col]) max_r = i;
if(a[max_r][col] == 0){
--k;
++free_num; continue;
}
if(max_r != k){
for(int i = col; i <= var; ++i)
swap(a[k][i], a[max_r][i]);
}
for(int i = k+1; i < equ; ++i) if(a[i][col])
for(int j = col; j <= var; ++j)
a[i][j] ^= a[k][j];
}
return var - k;
}
int main(){
int T; cin >> T;
while(T--){
scanf("%d", &n);
for(int i = 0; i < n; ++i){
scanf("%d", &m);
for(int j = 0; j < 31; ++j) a[j][i] = (m&(1<<j));
}
for(int i = 0; i < 31; ++i) a[i][n] = 0;
equ = 31; var = n;
int t = Gauss();
int ans = 1;
for(int i = 0; i < t; ++i)
ans = ans * 2 % mod;
printf("%d
", ans);
}
return 0;
}