Codeforces Round #142 (Div. 2)B. T-primes
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer tТ-prime, if t has exactly three distinct positive divisors.
You are given an array of n positive integers. For each of them determine whether it is Т-prime or not.
The first line contains a single positive integer, n (1 ≤ n ≤ 105), showing how many numbers are in the array. The next line contains nspace-separated integers xi (1 ≤ xi ≤ 1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64dspecifier.
Print n lines: the i-th line should contain "YES" (without the quotes), if number xi is Т-prime, and "NO" (without the quotes), if it isn't.
3
4 5 6
YES
NO
NO
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".
猜测 题。一猜 他是平方数 发现16不行 二猜 sqrt(16)应该是素数才行,再看一看诗句范围 正好。
/* *********************************************** Author :guanjun Created Time :2016/9/8 17:35:45 File Name :cf142b.cpp ************************************************ */ #include <iostream> #include <cstring> #include <cstdlib> #include <stdio.h> #include <algorithm> #include <vector> #include <queue > #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <iomanip> #include <list> #include <deque> #include <stack> #define ull unsigned long long #define ll long long #define mod 90001 #define INF 0x3f3f3f3f #define maxn 100010 #define cle(a) memset(a,0,sizeof(a)) const ull inf = 1LL << 61; const double eps=1e-5; using namespace std; priority_queue<int,vector<int>,greater<int> >pq; struct Node{ int x,y; }; struct cmp{ bool operator()(Node a,Node b){ if(a.x==b.x) return a.y> b.y; return a.x>b.x; } }; bool cmp(int a,int b){ return a>b; } ll a[maxn]; bool is(int x){ for(int i=2;i*i<=x;i++){ if(x%i==0)return false; } return true; } int main() { #ifndef ONLINE_JUDGE //freopen("in.txt","r",stdin); #endif //freopen("out.txt","w",stdout); int n; while(cin>>n){ for(int i=1;i<=n;i++)cin>>a[i]; for(int i=1;i<=n;i++){ if(a[i]==1){ puts("NO");continue; } ll x=sqrt(a[i]); if(x*x==a[i]&&is(x)){ puts("YES"); } else puts("NO"); } } return 0; }
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