HDU5833

每个数的质因子的大小不超过2000，2000以内质数大概300个。因为要选取若干数，使得他们乘起来是完全平方数，等价于把他们质因数的质数加起来，各质因数质数%2为0

考虑每个ai都是一个列向量，每个ai,j就是第i个数的第j个质因子的次数，那么要求的就是对于每一行都有∑0≤j<nai,j≡0 mod 2高斯消元搞一搞，数数自由元即可

////  Created by Running Photon//  Copyright (c) 2015 Running Photon. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <sstream>#include <set>#include <vector>#include <stack>#define ALL(x) x.begin(), x.end()#define INS(x) inserter(x, x,begin())#define ll long long#define CLR(x) memset(x, 0, sizeof x)using namespace std;const int inf = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int maxn = 1e5 + 10;const int maxv = 2e3 + 10;const double eps = 1e-9;int maxprime[maxv];std::vector <int> primes;int mp[311][311];void init() {    for(int i = 2; i <= 2e3; i++) {        if(!maxprime[i]) {            primes.push_back(i);            for(int j = i; j <= 2e3; j += i)                maxprime[j] = i;        }    }//  for(int i = 0; i < primes.size(); i++) {//      printf("%d\n", primes[i]);//  }}ll Pow(ll a, ll n) {    ll ret = 1;    while(n) {        if(n & 1) ret = ret * a % MOD;        a = a * a % MOD;        n >>= 1;    }    return ret;}void print(int n, int m) {    for(int i = 0; i < n; i++) {        for(int j = 0; j <= m; j++) {            printf("%d ", mp[i][j]);        }        puts("");    }}ll Gauss(int equ, int var) {//  print(equ, var);    int row = 0, col = 0;    int Free = 0;    for(row = 0, col = 0; row < equ && col < var; row++, col++) {        int maxRow = row;        for(int i = row + 1; i < equ; i++) {            if(mp[i][col] > mp[maxRow][col]) maxRow = i;        }        if(!mp[maxRow][col]) {            Free++;            row--;            continue;        }        if(maxRow != row) {            for(int j = col; j <= var; j++) swap(mp[row][j], mp[maxRow][j]);        }        for(int i = row + 1; i < equ; i++) {            if(mp[i][col]) {                for(int j = col; j <= var; j++) {                    mp[i][j] ^= mp[row][j];                }            }        }    }    return Pow(2, Free);}int main() {#ifdef LOCAL    freopen("C:\\Users\\Administrator\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);#endif//  ios_base::sync_with_stdio(0);    init();    int cas = 0, T;    scanf("%d", &T);    while(T--) {        printf("Case #%d:\n", ++cas);        int n = 0, m;        scanf("%d", &m);        CLR(mp);        for(int j = 0; j < m; j++) {            ll x;            scanf("%lld", &x);            int i = 0;            for(i = 0; x != 1; i++) {                int cnt = 0;                while(x % primes[i] == 0) {                    x /= primes[i];                    cnt++;                }                mp[i][j] = cnt % 2;            }            n = max(n, i);        }//      printf("n = %d  m = %d\n", n, m);        for(int i = 0; i < n; i++) {            mp[i][m] = 0;        }        printf("%lld\n", (Gauss(max(n, m), m) - 1 + MOD) % MOD);    }    return 0;}