Square Number:
Description
In mathematics, a square number is an integer that is the square of an integer. In other words, it is the product of some integer with itself. For example, 9 is a square number, since it can be written as 3 * 3.
Given an array of distinct integers (a1, a2, ..., an), you need to find the number of pairs (ai, aj) that satisfy (ai * aj) is a square number.
Input
The first line of the input contains an integer T (1 ≤ T ≤ 20) which means the number of test cases.
Then T lines follow, each line starts with a number N (1 ≤ N ≤ 100000), then N integers followed (all the integers are between 1 and 1000000).
Output
For each test case, you should output the answer of each case.
Sample Input
1 5 1 2 3 4 12
Sample Output
2
Cube Number:
Description
In mathematics, a cube number is an integer that is the cube of an integer. In other words, it is the product of some integer with itself twice. For example, 27 is a cube number, since it can be written as 3 * 3 * 3.
Given an array of distinct integers (a1, a2, ..., an), you need to find the number of pairs (ai, aj) that satisfy (ai * aj) is a cube number.
Input
The first line of the input contains an integer T (1 ≤ T ≤ 20) which means the number of test cases.
Then T lines follow, each line starts with a number N (1 ≤ N ≤ 100000), then N integers followed (all the integers are between 1 and 1000000).
Output
For each test case, you should output the answer of each case.
Sample Input
1 5 1 2 3 4 9
Sample Output
2
题意：
给你一列数，问两个数相乘组成（平方数&立方数）的种数有多少
题解：
对于平方数来说，每个平方数都能分解成若干质数的平方，所以枚举所有的素数，如果出现偶次幂直接忽略，若是奇数次幂，打表统计；
对于立方数来说，每个平方数都能分解成若干质数的立方，枚举所有的素数，若出现三次幂忽略，然后剩下的有两种情况：
例如剩下的一个数可以分解成三个质数a*b^2(a,b均为质数),那么他只能和a^2*b匹配；
代码（square number）：
#include <bits/stdc++.h> #include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <cstring> #include <vector> #include <map> #include <set> #include <bitset> #include <queue> #include <deque> #include <stack> #include <iomanip> #include <cstdlib> using namespace std; #define is_lower(c) (c>=‘a‘ && c<=‘z‘) #define is_upper(c) (c>=‘A‘ && c<=‘Z‘) #define is_alpha(c) (is_lower(c) || is_upper(c)) #define is_digit(c) (c>=‘0‘ && c<=‘9‘) #define min(a,b) ((a)<(b)?(a):(b)) #define max(a,b) ((a)>(b)?(a):(b)) #define IO ios::sync_with_stdio(0);\ cin.tie(0); cout.tie(0); #define For(i,a,b) for(int i = a; i <= b; i++) typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; typedef pair<ll,ll> pll; typedef vector<int> vi; const ll inf=0x3f3f3f3f; const double EPS=1e-10; const ll inf_ll=(ll)1e18; const ll mod=1000000007LL; const int maxn=1000000; bool vis[maxn+1000000]; int prime[maxn],prime1[maxn]; bool isprime[maxn+5]; int cnt[maxn+5]; int num = 0; void getprime() { memset(vis, false, sizeof(vis)); int N = sqrt(maxn); for (int i = 2; i <= N; ++i) { if ( !vis[i] ) { prime[++num] = i; prime1[num] = i*i; } for (int j = 1; j <= num && i * prime[j] <= N ; j++) { vis[ i * prime[j] ] = true; if (i % prime[j] == 0) break; } } } int main() { int T; cin>>T; getprime(); while(T--) { int x; memset(cnt,0,sizeof(cnt)); cin>>x; For(i,1,x) { int xx; cin>>xx; for(int j = 1; xx>=prime1[j]&&j<=num; j++) { while(xx%prime1[j]==0) xx/=prime1[j]; } cnt[xx]++; } ll ans = 0; For(i,1,maxn-1) if(cnt[i]) ans+= cnt[i]*(cnt[i]-1)/2; cout<<ans<<endl; } return 0; }
Cube Number:
#include <bits/stdc++.h> #include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <cstring> #include <vector> #include <map> #include <set> #include <bitset> #include <queue> #include <deque> #include <stack> #include <iomanip> #include <cstdlib> using namespace std; #define is_lower(c) (c>=‘a‘ && c<=‘z‘) #define is_upper(c) (c>=‘A‘ && c<=‘Z‘) #define is_alpha(c) (is_lower(c) || is_upper(c)) #define is_digit(c) (c>=‘0‘ && c<=‘9‘) #define min(a,b) ((a)<(b)?(a):(b)) #define max(a,b) ((a)>(b)?(a):(b)) #define IO ios::sync_with_stdio(0);\ cin.tie(0); cout.tie(0); #define For(i,a,b) for(int i = a; i <= b; i++) typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; typedef pair<ll,ll> pll; typedef vector<int> vi; const ll inf=0x3f3f3f3f; const double EPS=1e-10; const ll inf_ll=(ll)1e18; const ll mod=1000000007LL; const int maxn=1000000; bool vis[maxn+1000000]; ll prime[maxn],prime_2[maxn],prime_3[maxn]; bool isprime[maxn+5]; ll cnt[maxn+5]; int num = 0; void getprime() { memset(vis, false, sizeof(vis)); int N = maxn; for (int i = 2; i <= N; ++i) { if ( !vis[i] ) { prime[++num] = i; prime_2[num] = i * i; prime_3[num] = i * i * i; } for (int j = 1; j <= num && i * prime[j] <= N ; j++) { vis[ i * prime[j] ] = true; if (i % prime[j] == 0) break; } } } int main() { int T; cin>>T; getprime(); while(T--) { int x; int res = 0; cin >> x; memset(cnt, 0, sizeof(cnt)); for(int i = 1,xx; i <= x && cin>>xx; i++) { for(int j = 1; j <= num && xx >= prime_3[j]; j++) if(xx % prime_3[j]==0) while(xx % prime_3[j] == 0) xx /= prime_3[j]; cnt[xx]++; if(xx == 1) { res += cnt[xx] - 1; continue; } int tem = 1; for(int j = 1; j <=num && xx >= prime_2[j]; j++) if(xx % prime_2[j] == 0) while( xx % prime_2[j] == 0) { xx /= prime_2[j]; tem *= prime_2[j]; } if(xx <= 1000) { // 大于1000的素数的平方一定不存在； xx = sqrt(tem) * xx * xx; // 和另一半匹配，一定不大于maxn 加条件判断； if(xx < maxn) res += cnt[xx]; } } cout << res << endl; } return 0; }