键盘输入一个整数n,该整数代表下面数据的行数.

#includevoidmain(){\x09intn,i,j,t,a[10000];\x09printf("请输入正整数N:");\x09scanf("%d",&n);\x09printf("请依次

import java.util.Scanner;  public class TestJT {  public&nbs

步长改为2,就是n+=2；偶数一定不是素数.还有,被检验的数不是素数的话肯定能被之前的素数整除.所以用之前的检测出的素数除被检测书就行了,然后判断一个素数之需除到它的一半就行了.我只是为了运行速度提出

#includeusingnamespacestd;intmain(){inti,n,s=0;coutn;s=0;for(i=1;i

#includevoidperm(int*data,intn,intcurr){if(curr==n-1){for(inti=0;iprintf("%d",data[i]);printf("\n");

#include#includevoidmain(){inti,j,n,number=1,a[30][30];printf("PleaseinputanumberN:");scanf("%d",&n)

楼主你好!根据你要求实现如下#include<stdio.h>int fun(int n){ if(n>0)return n*fun(n-1)

/>#include<stdio.h>voidmain(){   intn,i,sum=1;   printf("in

#includeintmain(){intnum;inta,b,c;scanf("%d",&num);a=num/100;b=(num%100)/10;c=num%10;printf("a=%d,b=

a*a=b假设你先循环2到a,发现b都不能整除,这时你在循环a+1到a*a就没有意义了,因为b=a*a,所以b/(a+x)是肯定小于a的,而2到a已经循环过了不是吗?不过一般代码里面都是循环2到b/2

#include<stdio.h> int getMaxNumber(int n)  { int k; 

#include#includevoidmain(){intn;float*t,*s,*v;scanf("%d",&n);t=(float*)malloc(sizeof(float)*n);s=(fl

#include#includevoidmain(){intn,s,j,i;printf("inputn(n

#include <stdio.h>int main(){\x09int i,N;\x09scanf("%d",&N);\x09i

/*#include"stdio.h"main(){intx;printf("请输入一个数字:");scanf("%d",&x);if(x%3==0&&x%5==0)printf("该数字能同时被3,

oolcalc(unsignedintn){if(n

ntmain(intargc,char*argv[]){charsound[10][5]={"yi","er","san","si","wu","liu","qi","ba","jiu","shi"}

main函数：Scannerinput=newScanner(System.in);intnum=input.nextInt();if(num%3==0||num%5==0){system.out.p

publicstaticvoidmain(String[]args){Scannerscan=newScanner(System.in);while(scan.hasNextInt()){i=scan

已通过测试,#includemain(){inta[20],length;inti,j,t;printf("Inputthelengthofarray:");scanf("%d",&length);f