下面的程序EnhancedForDemo使用增强for循环遍历这个数组: public class EnhancedForDemo { public static void main(String[] args) { int[] numbers = {1,2,3,4,5,6,7,8,9,10}; for(int item : numbers){ System.out.println(Countis:+item); } } }在这个例子中,变量item保存从数字数组获得的当前...
include<bits/stdc++.h> using namespace std;int n,a[21][21],k;int main(){ cin>>n;for(int i=0;i<n;i++)for(int j=0;j<n;j++){ a[i][j]=k++;} for(int i=0;i<n;i++){ for(int j=0;j<n;j++)cout<<a[i][j]<<" ";cout<<endl;} return 0;} for...
PrintMatrix((int *)matrixA,4);printf("After convertion:\n");ConvertMatrix((int *)matrixB,(int *)matrixA,4);PrintMatrix((int *)matrixB,4);return 0;} int PrintMatrix(int *matrix,int n){ int i,j;if(!matrix || n<1)return 0;for(i=0;i<n;++i){ for(j=0;j<n...
B。int a[3][2]={1, 2, 3, 4, 5, 6}；定义a为3*2（3行2列）的数组，有6个元素。该数组的下标变量共有3×2个，即：a[0][0],a[0][1]a[1][0],a[1][1]a[2][0],a[2][1]数组中的每个元素都由元素名以[i，j]的形式标识，其中a是数组名，i和j是唯一标识a中每个...
public static void main(String[] args) {int[][] a = new int[4][5];int[] b = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19};int k = b.length - 1;for(int i = a.length - 1; i >= 0; i--) {for(int j = a[i].length - 1; j >= 0...
i应该从2开始include<stdio.h>void main(){int i=2,j=1;while(i<=10){ j*=i;i++;}printf("%d\n",j);}include <stdio.h>void main(){int a,s;s=1;for(a=1;a<=10;a++)s=s*i;printf("%d\n",s);}include<stdio.h>void main(0{int i=1,j=1;while(i<=10){j*...
for (int i = 0; i < 5; i++){ for (int j = 0; j < 5; j++){ int nPos = (i * 4 + j) % 5;printf("%d", a[nPos]);} printf("\n");}
include <stdio.h> int main(void){ for(int i=1;i<=6;i++){ for(int j=1;j<=i;j++){ printf("%d",j);} printf("\n");} } include
解答如下：当n=1时，排列为1 2，逆序数t=0。当n=2时，排列为内1 3 2 4，逆序容数t=1。当n=3时，排列为1 3 5 2 4 6，逆序数t=1+2=3。当n=4时，排列为1 3 5 7 2 4 6 8，逆序数t=1+2+3=6。当n=5时，排列为1 3 5 7 9 2 4 6 8 10，逆序数t=1+2+3+4=10...
现在看太冗长了些，不过至少是对的╮(╯_╰)╭这个是C++，你要C语言的话把里面的cin改成scanf,把cout改成printf，头文件的<iostream>改成<stdio.h>就好include<iostream>#include<math.h>using namespace std;int TentoTwo(int n){int b[10000],a[10000],i,j;for(i=0;n>=2;i++){a[...