Square
Input: Standard Input
Output: Standard Output
Given n integers you can generate 2n-1 non-empty subsets from them. Determine for how many of these subsets the product of all the integers in that is a perfect square. For example for the set {4,6,10,15} there are 3 such subsets. {4}, {6,10,15} and {4,6,10,15}. A perfect square is an integer whose square root is an integer. For example 1, 4, 9,16,…. .
Input
Input contains multiple test cases. First line of the input contains T(1≤T≤30) the number of test cases. Each test case consists of 2 lines. First line contains n(1≤n≤100) and second line contains n space separated integers. All these integers are between 1 and 10^15. None of these integers is divisible by a prime greater than 500.
Output
For each test case output is a single line containing one integer denoting the number of non-empty subsets whose integer product is a perfect square. The input will be such that the result will always fit into signed 64 bit integer.
Sample Input Output for Sample Input
|
4 3 2 3 5 3 6 10 15 4 4 6 10 15 3 2 2 2 |
0 1 3 3 |
Problemsetter: Abdullah al Mahmud
Special Thanks to: Manzurur Rahman Khan
题目大意：给n个整数，从中选1个或多个，他们的积是一个完全平方数。这样的情况有多少种？
解题思路：选s个整数出来(1<=s<=n),它们的积可以表示成a1^p1*a2.P2....an^pm,要想它是一个完全平方数，那pi必须是偶数。把每个数分解质因数，所有质因数的指数都模2，最后得出一个n*m的矩阵，进行消元，求出矩阵的秩r。那么就有n-r行每个元素都是0。从中选一行或多行（即求一个集合的真子集个数）都符合题目要求。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 6 typedef long long LL; 7 const int maxn=500; 8 const int maxm=105; 9 typedef int Matrix[maxm][maxm]; 10 int prime[maxn],num; 11 bool flag[maxn]; 12 Matrix A; 13 14 int max(int a,int b){ return a>b?a:b;} 15 void swap(int& a,int& b){int t=a;a=b;b=t;} 16 17 void get_primes() 18 { 19 int i,j;num=0; 20 memset(flag,1,sizeof(flag)); 21 for(i=2;i<maxn;i++) 22 { 23 if(flag[i])prime[num++]=i; 24 for(j=0;j<num&&i*prime[j]<maxn;j++) 25 { 26 flag[i*prime[j]]=false; 27 if(i%prime[j]==0) break; 28 } 29 } 30 } 31 32 int rank(int n,int m) 33 { 34 int i=0,j=0,k,r,u; 35 while(i<n&&j<m) 36 { 37 r=i; 38 for(k=i;k<n;k++) 39 if(A[k][j]){r=k;break;} 40 if(A[r][j]) 41 { 42 if(r!=i) for(k=0;k<=m;k++) swap(A[r][k],A[i][k]); 43 for(u=i+1;u<n;u++) if(A[u][j]) 44 for(k=i;k<=m;k++) A[u][k]^=A[i][k]; 45 i++; 46 } 47 j++; 48 } 49 return i; 50 } 51 52 LL Pow(LL a,LL b) 53 { 54 LL ret=1; 55 while(b) 56 { 57 if(b&1) ret*=a; 58 a*=a;b>>=1; 59 } 60 return ret; 61 } 62 63 int main() 64 { 65 get_primes(); 66 int i,j,t,n,maxp; 67 LL p; 68 scanf("%d",&t); 69 while(t--) 70 { 71 scanf("%d",&n); 72 memset(A,0,sizeof(A)); 73 maxp=0; 74 for(i=0;i<n;i++) 75 { 76 scanf("%lld",&p); 77 for(j=0;j<num;j++) 78 { 79 if(p==1) break; 80 while(p%prime[j]==0) 81 { 82 maxp=max(maxp,j); 83 p/=prime[j]; 84 A[i][j]^=1; 85 } 86 } 87 } 88 int r=rank(n,maxp+1); 89 printf("%lld ",Pow(2,n-r)-1); 90 } 91 return 0; 92 }