c++编程T-primes number

c++编程T-primes number
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 ≤ 10000),showing how many numbers are in the array.
The next line contains n space-separated integers xi (1 ≤ xi ≤ 10^12).
Output
Print one line:The number of T-primes number
Time Limit:1sec! 遗忘的大海 1年前他留下的回答 已收到1个回答

牙牙2000 网友

该名网友总共回答了20个问题，此问答他的回答如下：采纳率：85%

//#include "stdafx.h"//vc++6.0加上这一行.
#include "stdio.h"
#include "stdlib.h"
bool T_prime(__int64 n){
x09 __int64 i,x;
x09 if(n

1年前他留下的回答 追问

8

遗忘的大海

貌似不对啊~~~~~~

牙牙2000

//#include "stdafx.h"//vc++6.0加上这一行.
#include "stdio.h"
#include "stdlib.h"
bool T_prime(__int64 n){
__int64 i,x;
if(n<4)x09return false;
for(i=2;(x=i*i) if(n%i==0) return false;
if(x==n) return true;
else return false;
}
void main(void){
int n,i;
__int64 *xi;
while(1){
printf("Type n(1<=n<=10000)...nn=");
scanf("%d",&n);
if(n>0 && n<10001) break;
printf("Error, redo: ");
}
if((xi=(__int64 *)malloc(n*sizeof(__int64)))==NULL){
printf("Application memory failure...n");
exit(0);
}
for(i=0;i printf("======================================n");
for(i=0;i if(T_prime(xi[i]))
printf("%I64d ",xi[i]);
free(xi);
printf("n");
}
说出不对来……

遗忘的大海

没错。。谢啦

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