UVa 11542 (高斯消元 异或方程组) Square

书上分析的太清楚，我都懒得写题解了。=_=||

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

const int maxp = 100;
const int maxn = 500;
bool vis[maxn + 10];
int prime[maxp], pcnt = 0;

void Init()
{
    int m = sqrt(maxn + 0.5);
    for(int i = 2; i <= m; i++) if(!vis[i])
        for(int j = i*i; j <= maxn; j += i) vis[j] = true;
    for(int i = 2; i <= maxn; i++) if(!vis[i]) prime[pcnt++] = i;
}

typedef int Matrix[maxn][maxn];

Matrix A;

int rank(Matrix A, int m, int n)
{//求系数矩阵A的秩，m个方程，n个未知数
    int i = 0, j = 0;
    while(i < m && j < n)
    {
        int r = i, k;
        for(k = r; k < m; k++) if(A[k][j]) { r = k; break; }
        if(k < m)
        {
            if(r != i) for(int k = 0; k < n; k++) swap(A[r][k], A[i][k]);
            for(int k = i+1; k < m; k++) if(A[k][j])
                for(int l = j; l < n; l++) A[k][l] ^= A[i][l];
            i++;
        }
        j++;
    }
    return i;
}

int main()
{
    //freopen("in.txt", "r", stdin);

    Init();
    int T;
    scanf("%d", &T);
    while(T--)
    {
        memset(A, 0, sizeof(A));
        int n, M = 0;
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
        {
            long long x;
            scanf("%lld", &x);
            for(int j = 0; j < pcnt; j++) while(x % prime[j] == 0)
            {
                M = max(M, j);
                x /= prime[j];
                A[j][i] ^= 1;
            }
        }
        int r = rank(A, M+1, n);//共用到前M+1个素数
        printf("%lld
", (1LL << (n-r)) - 1);
    }

    return 0;
}

最后lrj老师提到了还可以用状压加速消元，因为500以内的素数不超过100个，所以我用了两个64位的long long来表示一个方程。第一份代码16ms，状压以后12ms，快了四分之一。

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

const int maxp = 100;
const int maxn = 500;
bool vis[maxn + 10];
int prime[maxp], pcnt = 0;

void Init()
{
    int m = sqrt(maxn + 0.5);
    for(int i = 2; i <= m; i++) if(!vis[i])
        for(int j = i*i; j <= maxn; j += i) vis[j] = true;
    for(int i = 2; i <= maxn; i++) if(!vis[i]) prime[pcnt++] = i;
}

typedef long long Matrix[maxn][2];

Matrix A;

int rank(Matrix A, int m, int n)
{//求系数矩阵A的秩，m个方程，n个未知数
    int i = 0, j = 0, len = n / 64;
    while(i < m && j < n)
    {
        int r = i, k;
        for(k = r; k < m; k++) if(A[k][j/64] & (1LL<<(j%64))) { r = k; break; }
        if(k < m)
        {
            if(r != i) for(int k = 0; k <= len; k++) swap(A[r][k], A[i][k]);
            for(int k = i+1; k < m; k++) if(A[k][j/64] & (1LL<<(j%64)))
                for(int l = 0; l <= len; l++) A[k][l] ^= A[i][l];
            i++;
        }
        j++;
    }
    return i;
}

int main()
{
    //freopen("in.txt", "r", stdin);

    Init();
    int T;
    scanf("%d", &T);
    while(T--)
    {
        memset(A, 0, sizeof(A));
        int n, M = 0;
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
        {
            long long x;
            scanf("%lld", &x);
            for(int j = 0; j < pcnt; j++) while(x % prime[j] == 0)
            {
                M = max(M, j);
                x /= prime[j];
                A[j][i/64] ^= (1LL << (i%64) );
            }
        }
        int r = rank(A, M+1, n);//共用到前M+1个素数
        printf("%lld
", (1LL << (n-r)) - 1);
    }

    return 0;
}