HDU 5833 Zhu and 772002 （高斯消元）

Zhu and 772002

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 396    Accepted Submission(s): 121

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 by1000000007.

 

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

233 3 432 2 2

 

Sample Output

Case #1:3Case #2:3

 

Author

UESTC

 

Source

2016中国大学生程序设计竞赛 - 网络选拔赛

和uva 11542 一样  看那个博客吧！
只是数据稍微大了一点，思路还是一样的！
听说学长在取模上有坑，可能是在乘的时候不小心爆了。。
因为要取模，所以可以写个快速幂，这样可以解决溢出问题！

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<cstdlib>#include<vector>using namespace std;typedef long long ll;const int maxn = 2000 + 5;const int mod = 1000000007;int equ, var;int a[maxn][maxn];int x[maxn];int free_x[maxn];int free_num;int vis[maxn];vector<int >prime;int len_prime; ll my_pow(ll a,ll n){    ll ans = 1;    while(n){        if (n & 1)            ans = (ans % mod * a % mod) % mod;        n/=2;        a = (a%mod*a%mod)%mod;    }    return ans; }int Gauss(){    int max_r,col,k;    free_num = 0;    for (k = 0, col = 0; k < equ && col < var; k++, col++){        max_r = k;        for (int i = k+1; i < equ; i++){            if (abs(a[i][col]) > abs(a[max_r][col]))                max_r = i;        }        if (a[max_r][col] == 0){            k--;            free_x[free_num++] = col;            continue;        }        if (max_r != k){            for (int j = col; j < var+1; j++)                swap(a[k][j],a[max_r][j]);        }        for (int i = k+1; i < equ; i++){            if (a[i][col] != 0){                for (int j = col; j < var+1; j++)                    a[i][j] ^= a[k][j];            }        }    }    for (int i = k; i < equ; i++)        if (a[i][col] != 0)            return -1;    if (k < var)return var- k;    for (int i = var-1; i >= 0; --i){        x[i] = a[i][var];        for (int j = i+1; j < var; j++)            x[i] ^= (a[i][j] && x[j]);    }    return 0;}void init(){    int len = sqrt(maxn) + 1;    for (int i = 2; i <= len; ++i) if (!vis[i])        for (int j = i*i; j < maxn; j+=i) vis[j] = 1;    for (int i = 2; i < maxn; ++i)if (!vis[i])prime.push_back(i);    len_prime = (int)prime.size();}int main(){    init();    int T,kase = 0;    scanf("%d",&T);    while(T--){        int n;        scanf("%d",&n);        memset(a,0,sizeof a);        for (int i = 0; i < n; ++i){            ll x;            scanf("%I64d",&x);            for (int j = 0; j < len_prime; ++j){                int pri = prime[j];                while(x % pri == 0){                    a[j][i] ^= 1;                    x/=pri;                }                if (x == 1)break;            }        }        equ = len_prime;        var = n;        int t = Gauss();        printf("Case #%d:\n",++kase);        printf("%I64d\n",my_pow(2ll,(ll)t)-1ll);    }    return 0;}