Zhu and 772002
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1799 Accepted Submission(s): 624

Problem Description
Zhu and 772002 are both good at math. One day, Zhu wants to test the ability of 772002, so he asks 772002 to solve a math problem.

But 772002 has a appointment with his girl friend. So 772002 gives this problem to you.

There are n numbers a1,a2,…,an. The value of the prime factors of each number does not exceed 2000, you can choose at least one number and multiply them, then you can get a number b.

How many different ways of choices can make b is a perfect square number. The answer maybe too large, so you should output the answer modulo by 1000000007.

Input
First line is a positive integer T , represents there are T test cases.

For each test case:

First line includes a number n(1≤n≤300)，next line there are n numbers a1,a2,…,an,(1≤ai≤1018).

Output
For the i-th test case , first output Case #i: in a single line.

Then output the answer of i-th test case modulo by 1000000007.

Sample Input

2
3
3 3 4
3
2 2 2

Sample Output

Case #1:
3
Case #2:
3

解题方法： 白书原题，看这位小哥的吧。见这里

//HDU 5833 XOR GUASS

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2010;
typedef long long LL;
typedef int Matrix[maxn][maxn];
const int mod = 1000000007;
Matrix A; //A[i][j]就表示第j个数的这个i素数是奇数还是偶数
int n, vis[2010], prime[2010];
int pre_deal(int m){
    memset(vis, 0, sizeof(vis));
    int cnt = 0;
    for(int i = 2; i < m; i++){
        if(!vis[i]){
            prime[cnt++] = i;
            for(int j = i*i; j < m; j += i){
                vis[j] = 1;
            }
        }
    }
    return cnt;
}
LL powmod(LL a, LL n){
    LL res = 1;
    while(n){
        if(n&1) res = res*a%mod;
        a = a*a%mod;
        n >>= 1;
    }
    return res;
}
int xor_guass(int m, int n) //A是异或方程组系数矩阵 返回秩
{
    int i = 0, j = 0, k, r, u;
    while(i < m && j < n){//当前正在处理第i个方程,第j个变量
        r = i;
        for(int k = i; k < m; k++) if(A[k][j]){r = k; break;}
        if(A[r][j]){
             if(r != i) for(k = 0; k <= n; k++) swap(A[r][k], A[i][k]);
             //消元完成之后第i行的第一个非0列是第j列,且第u>i行的第j列全是0
            for(u = i + 1; u < m; u++) if(A[u][j])
                for(k = i; k <= n; k++) A[u][k] ^= A[i][k];
            i++;
        }
        j++;
    }
    return i;
}
int main()
{
    int tot = pre_deal(maxn);
    int T, ks = 0; scanf("%d", &T); while(T--){
        memset(A, 0, sizeof(A));
        scanf("%d", &n);
        int maxp = 0;
        for(int i = 0; i < n; i++){
            LL x;
            scanf("%lld", &x);
            for(int j = 0; j < tot; j++){
                while(x % prime[j] == 0){
                    maxp = max(maxp, j);
                    x /= prime[j];
                    A[j][i] ^= 1;
                }
            }
        }
        int rr = xor_guass(maxp + 1, n);//秩
        printf("Case #%d:\n", ++ks);
        printf("%lld\n", powmod(2, n - rr) - 1);
    }
    return 0;
}

当然我们不能这样A了就算了，在大白树上提到了一种bitset优化的方法，即是bitset压位优化。即是把32列合并到一个无符号32位整数中，然后只需要用一次逐位异或xor就可以处理32列了，所以复杂度可以降到O(n*n*n/32)，扣扣常数还是非常支持的。

//HDU 5833 XOR GUASS bitset

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2005;
const int mod = 1e9+7;
int n, vis[maxn], prime[maxn], cnt;
void pre_deal(){
    for(int i = 2; i < maxn; i++){
        if(vis[i]) continue;
        prime[cnt++] = i;
        for(int j = i; j < maxn; j += i) vis[j] = 1;
    }
}
bitset <330> A[305]; //A[i][j]就表示第j个数的这个i素数是奇数还是偶数

int main()
{
    pre_deal();
    int T, ks = 0; scanf("%d", &T);
    while(T--){
        printf("Case #%d:\n", ++ks);
        for(int i = 0; i < 305; i++) A[i].reset();
        scanf("%d", &n);
        for(int i = 0; i < n; i++){
            long long x;
            scanf("%lld", &x);
            for(int j = 0; j < cnt; j++){
                if(x%prime[j] == 0){
                    int flag = 0;
                    while(x%prime[j] == 0){
                        x /= prime[j];
                        flag ^= 1;
                    }
                    A[j][i] = flag;
                }
            }
        }
        int i = 0, j = 0; //xor消元之后j就是秩
        for(i = 0; i < n; i++){
            int id = -1;
            for(int k = j; k < cnt; k++){
                if(A[k][i]){
                    id = k;
                    break;
                }
            }
            if(id == -1) continue;
            swap(A[j], A[id]);
            for(int k = j + 1; k < cnt; k++){
                if(A[k][i]) A[k] ^= A[j];
            }
            j++;
        }
        int ans = 1;
        for(int i = 0; i < n - j; i++) ans = ans * 2 % mod;
        ans--;
        printf("%d\n", ans);
    }
    return 0;
}

所以一下午只写了2个高斯消元？？效率需要挺高啦。