求编一个小程序C++：判断一个数是否是质数，考虑异常情况。

#include<iostream>

using namespace std

int judge(int n)//判断一个自然数是否为质数

{

for(int i=2i<=n/2i++)

if(n%i==0)

return 0

return 1

}

void main()

{

float n

cout<<"输入一个数字：丛启"<<endl

cin>>n

int m

m=(int)(n)

try

{

if(n<=0||n!=m) throw m//渗庆如异常检测差裂，判断否为负数或者不为整数

if(n>0&&n==m) throw "ok"

}

catch(int)

{

cout<<"数字不合法"<<endl

}

catch(char *)

{

if(judge(n))

cout<<"素数"<<endl

else

cout<<"非素数"<<endl

}

}

PS：其实ls说的很对

if(文本框1.getText().equals(文本框2.getText())) System.out.println("两个文本框内容相等")//或者 if(!文本框纯芹睁1.getText().equals(文本框2.getText())) System.out.println("两首缺个文本框内容不相等"做岁)

挺简单的啊。。 关键在于理解迟宽heap

import java.util.Arrays

public class Du {

public static void main(String[] args) {

int[] testAry1 = {1, 3, 5, 7, 9, 11}

System.out.println("{1, 3, 5, 7, 9, 11} is " + (isHeap(testAry1)? "": "NOT ") + "heap array.")

int[] testAry2 = {1, 3, 5, 7, 13, 9, 11}

System.out.println("枝隐{1, 3, 5, 7, 13, 9, 11} is "猛旦厅 + (isHeap(testAry2)? "":"NOT ") + "heap array.")

}

/**

* Will receive an integer heap array, method will return true or false

* depending on whether the array is believed to be a valid heap or not

*

* @param heap

*0-based integer array

* @return true if integer array is a heap, else false

*/

public static boolean isHeap(int[] heap) {

if(heap == null || heap.length == 0){

return false

}

int[] ary = new int[heap.length]

for(int i = 0i <heap.lengthi++){

ary[i] = heap[i]

}

Arrays.sort(ary)

for(int i = 0i <heap.lengthi++){

if(ary[i] != heap[i]){

return false

}

}

return true

}

}

---------------

{1, 3, 5, 7, 9, 11} is heap array

{1, 3, 5, 7, 13, 9, 11} is NOT heap array

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