【POJ 1830】 开关问题

【题目链接】

http://poj.org/problem?id=1830

【算法】

列出异或方程组，用高斯消元求解

【代码】

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;

int i,j,k,T,n,x,y,ans;
int a[];

int main()
{

        scanf("%d",&T);
        while (T--)
        {
                scanf("%d",&n);
                for (i = ; i <= n; i++) scanf("%d",&a[i]);
                for (i = ; i <= n; i++)
                {
                        scanf("%d",&x);
                        a[i] ^= x;
                        a[i] |= ( << i);
                }
                while (scanf("%d%d",&x,&y) && x && y) a[y] |= ( << x);
                ans = ;
                for (i = ; i <= n; i++)
                {
                        for (j = i + ; j <= n; j++)
                        {
                                if (a[j] > a[i])
                                        swap(a[i],a[j]);
                        }
                        if (a[i] == )
                        {
                                ans =  << (n - i + );
                                break;
                        }
                        if (a[i] == )
                        {
                                ans = ;
                                break;
                        }
                        for (k = n; k; k--)
                        {
                                if (a[i] & ( << k))
                                {
                                        for (j = ; j <= n; j++)
                                        {
                                                if (i != j && (a[j] & ( << k)))
                                                        a[j] ^= a[i];
                                        }
                                        break;
                                }
                        }
                }
                if (!ans) printf("Oh,it's impossible~!!\n");
                else printf("%d\n",ans);
        }

        return ;

}