HDU 5833 Zhu and 772002

高斯消元。求有多少不同组解。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-8;
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar(); x = 0;while(!isdigit(c)) c = getchar();
    while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar();  }
}

const int MAXN=310;
int equ, var;///equ个方程 var个变量
int a[MAXN][MAXN];///增广矩阵
int x[MAXN];///解的数目
bool free_x[MAXN];///判断是不是自由变元
int free_num;///自由变元的个数

inline int GCD(int m, int n)
{
    if(n == 0)
        return m;
    return GCD(n, m%n);
}
inline int LCM(int a, int b)
{
    return a/GCD(a,b)*b;
}

int Gauss()
{
    int Max_r;///当前列绝对值最大的存在的行
    ///col：处理当前的列
    int row = 0;
    int free_x_num;
    int free_index = -1;
    for(int col=0; row<equ&&col<var; row++,col++)
    {
        Max_r = row;
        for(int i=row+1; i<equ; i++)
            if(abs(a[i][col]) > abs(a[Max_r][col]))
                Max_r = i;

        if(Max_r != row)
            for(int i=0; i<var+1; i++)
                swap(a[row][i], a[Max_r][i]);

        if(a[row][col] == 0)
        {
            row--;
            continue;
        }
        for(int i=row+1; i<equ; i++)
        {
            if(a[i][col])
            {
                int lcm = LCM(abs(a[i][col]), abs(a[row][col]));
                int tp1=lcm/abs(a[i][col]), tp2=lcm/abs(a[row][col]);
                if(a[row][col]*a[i][col] < 0)
                    tp2 = -tp2;
                for(int j=col; j<var+1; j++)
                {
                    a[i][j] = tp1*a[i][j]-tp2*a[row][j];
                    a[i][j] = (a[i][j]%2+2)%2;
                }
            }
        }
    }
    if(row < var)
    {
        for(int i=row-1; i>=0; i--)
        {
            free_x_num = 0;
            for(int j=0; j<var; j++)
                if(a[i][j] && free_x[j])
                {
                    free_x_num++;
                    free_index = j;
                }
            if(free_x_num>1 || free_index==-1)
                continue;
            int tmp = a[i][var];
            for(int j=0; j<var; j++)
                if(a[i][j] && j!=free_index)
                {
                    tmp -= a[i][j]*x[j];
                    tmp = (tmp%2+2)%2;
                }
            x[free_index] = (tmp*a[i][free_index]);/// 求出该变元.
            x[free_index] %= 2;
            free_x[free_index] = 0; /// 该变元是确定的.
        }
        return var - row;///自由变元的个数
    }

    for(int i=var-1; i>=0; i--)
    {
        int tmp = a[i][var];
        for(int j=i+1; j<var; j++)
            if (a[i][j])
            {
                tmp -= a[i][j]*x[j];
                tmp = (tmp%2+2)%2;
            }
        x[i] = tmp*a[i][i];
        x[i] %= 2;
    }
    return 0;///唯一解
}

const int maxn=2005;
bool f[maxn]; int p[maxn], t;

void prime()
{
    for(int i=2;i<=2000;i++) f[i]=1;
    for(int i=2;i<=2000;i++)
    {
        if(!f[i]) continue; p[t++]=i;
        int tmp=2*i; while(tmp<=2000) f[tmp]=0,tmp=tmp+i;
    }
}

int T,n;

int main()
{
    prime(); int cas=1;
    scanf("%d",&T); while(T--)
    {
        scanf("%d",&n);

        memset(a,0,sizeof a);
        memset(x,0,sizeof x);
        memset(free_x,0,sizeof free_x);
        free_num=0;

        for(int i=0;i<n;i++)
        {
            LL num; scanf("%lld",&num);
            for(int j=0;j<t;j++)
            {
                while(num%p[j]==0)  num=num/p[j],a[j][i]++;
                a[j][i]=a[j][i]%2;
            }
            
        }

        equ=t; var=n;
        int S = Gauss();

        long long ans=1, MOD=1e9+7;
        for(int i=1;i<=S;i++) ans=ans*2%MOD;
        printf("Case #%d:\n",cas++);
        printf("%lld\n",(ans-1+MOD)%MOD);

    }
    return 0;
}