1034: T-primes
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 1691 Solved: 316
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Description
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer t Т-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.
Input
The first line contains a single positive integer, n (1 ≤ n ≤ 105), showing how many numbers are in the array. The next line contains n space-separated integers xi (1 ≤ xi ≤ 109).
Output
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.
Sample Input
3
4 5 6
Sample Output
YES NO NO
得:
- 要換一種思路,想想會不會有更簡單的算法;要巧妙一點
分析:
- 一開始是定義一個函數,然後對輸入的每個數進行代入函數,計算一共有幾個除數,但是會提示數組越界、時間超限、答案錯誤。。。我也很無奈的。。。
- 然後,看了網上的代碼,不用定義數組,因為隻輸入一組數,循環然後一個個輸入就好;
- 不用定義函數,如果是T素數的話,就一定是素數平方得到的數;定義一個變量獲得輸入數的開方值;然後循環,j*j<這個值,即求該數是否為素數,如果是并且b的平方==a(有的數開方後是小數),則就是t素數;
代碼:
#include<bits/stdc+.h>
using namespace std;
int main()
{
int n,i,j;
long long a,b;
scanf("%d",&n);
for(i=1; i<=n; i++)
{
scanf("%lld",&a);
b = sqrt(a);
for(j=2; j*j<=b; j++)
{
if( b%j==0 )
break;
}
if(j*j>b && b*b==a && a>1)
{
printf("YES\n");
}else
{
printf("NO\n");
}
}
return 0;
}