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 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   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; }