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".
题意:判断一个数是否只有3个因子
思路:该数一定是质数的平方。
# include <stdio.h> # include <math.h> # define MAXN 1000000 int p[MAXN+1]={1,1}; int main() { int i, j, t; long long n; double temp; for(i=2; i<=MAXN; ++i) if(!p[i]) for(j=i+i; j<=MAXN; j+=i) p[j] = 1; scanf("%d",&t); while(t--) { scanf("%I64d", &n); temp = sqrt(n); if(temp == (int)temp && !p[(int)temp]) puts("YES"); else puts("NO"); } return 0; }