hdu 3915 Game 求N个数中取若干个数字使得它们的异或值为0的方法数高斯消元(mod2)

该博客介绍了如何使用高斯消元法解决一个数学问题，即在给定的一组整数中找出若干个数字，使得它们的异或值为0。博主通过编写C++代码实现了一个名为Gauss_XOR的函数，该函数接受矩阵和变量数作为参数，并返回解的数量。在主函数中，博主读取输入数据，应用高斯消元并输出结果。如果无法找到异或值为0的组合，则输出-1。

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Problem Description

Mr.Frost is a child who is too simple, sometimes naive, always plays some simple but interesting games with his friends. Today，he invents a new game again:
At the beginning of the game they pick N (1<=N<=100) piles of stones, Mr.Frost and his friend move the stones in turn. At each step of the game, the player chooses a pile, removes at least one stone from the pile, the first player can’t make a move, and loses. So smart is the friends of Mr.Frost that Mr.Frost always loses. Having been a loser for too many times, he wants to play a trick. His plan is to remove some piles, and then he can find a way to make sure that he would be the winner after his friends remove stones first.

Now, he wants to know how many ways to remove piles which are able to achieve his purpose. If it’s impossible to find any way, please print “-1”.

Input

The first line contains a single integer t (1<=t<=20), that indicates the number of test cases. Then follow the t cases. Each test case begins with a line contains an integer N (1 <= N <= 100), representing the number of the piles. The next n lines, each of which has a positive integer Ai(1<=Ai<=2^31 - 1) represent the number of stones in this pile.

Output

For each case, output a line contains the number of the way mod 1000007, If it’s impossible to find any way, please print “-1”.

Sample Input

Sample Output

2
2

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define maxn 110
using namespace std;

typedef int Int;

int T, N;
int beg[maxn], end[maxn], x[maxn];
int a[maxn][maxn];

int Gauss_XOR(Int a[maxn][maxn], Int x[maxn], int var, int equ)
{
    int row, col;
    for (row = col = 1; row <= equ && col <= var; ++row, ++col)
    {
        if (!a[row][col])
        {
            for (int i = equ; i > row; --i)
            {
                if (a[i][col])
                {
                    for (int j = row; j <= var + 1; ++j)
                    {
                        swap(a[i][j], a[row][j]);
                    }
                    break;
                }
            }
        }
        if (!a[row][col])
        {
            --row;
            continue;
        }
        for (int i = row + 1; i <= equ; ++i)
        {
            if (a[i][col])
            {
                for (int j = var + 1; j >= col; --j)
                {
                    a[i][j] ^= a[row][j];
                }
            }
        }
    }
    for (int i = row; i <= equ; ++i)
    {
        if (a[i][var + 1]) return -1;
    }
    if (row <= var)
    {
        return var - row + 1;
    }
    for (int i = var; i >= 1; --i)
    {
        x[i] = a[i][var + 1];
        for (int j = i + 1; j <= var; ++j)
        {
            x[i] ^= a[i][j] && x[j];
        }
    }
    return 0;
}

int main()
{
    int num;
    scanf("%d", &T);
    while (T--)
    {
        int equ = 0;
        memset(x, 0, sizeof (x));
        memset(a, 0, sizeof (a));
        scanf("%d", &N);
        for(int i = 1; i <= N; ++i)
        {
            int pos = 1;
            scanf("%d", &num);
            while(num)
            {
                if(num & 1) a[pos][i] = 1;
                else a[pos][i] = 0;
                num >>= 1;
                ++pos;
            }
            equ = max(equ, pos - 1);
        }
        for(int i = 1; i <= 32; ++i)
            a[i][N + 1] = 0;
        int ans = Gauss_XOR(a, x, N, equ);
        if (ans == -1) puts("-1");
        else
  {
   int prt = 1;
   for(int i = 1; i <= ans; ++i)
   {
    prt <<= 1;
    prt %= 1000007;
   }
   printf("%d\n", prt);
  }
    }
    return 0;
}