、从键盘输入一个3行3列的二维整型数组(共9个整数元素)。

include <stdio.h>int main(){ int a[3][3]; int i,j; int max; //1 for(i = 0; i<3;i++) for(j = 0; j < 3; j ++) scanf("%d",&a[i][j]); //2 for(i = 0; i<3;i++) { for(j = 0; j < 3; j ++) printf("...

include "stdio.h"int main(int argc,char *argv[]){int a[3][3],s,i,j;printf("Please enter data...\n");for(i=0;i<3;i++)//输入3x3数组for(j=0;j<3;scanf("%d",a[i]+j++));printf("\n");for(s=i=0;i<3;i++){//按3行3列输出并求对角线之和for(j=0;j<...

//#include "stdafx.h"//vc++6.0加上这一行.#include "stdio.h"int main(void){ int a[3][3],sum,i,j; for(sum=i=0;i<3;i++) for(j=0;j<3;j++){ scanf("%d",a[i]+j); sum+=a[i][j]; } printf("The result is %d\n",sum); return ...

include <stdio.h>const int N = 3;void Transpose(int *a,int n) {int i,j,t;for(i = 0; i < n - 1; ++i) {for(j = i + 1; j < n; ++j) {t = *(a + n * i + j);*(a + n * i + j) = *(a + n * j + i);*(a + n * j + i) = t;}}}...

void main(){ void change(int x[][3],int y);int x[3][3]={{1,2,3},{4,5,6},{7,8,9}};int i,j;printf("原数组为\n");for(i=0;i<=2;i++){ for(j=0;j<=2;j++){ printf("%3d",x[i][j]);} printf("\n");} change(x,3);printf("行列互换后的数组为...

int main(void){ int a[3][3];int i = 0, j = 0,num=0;printf("输入一个整型二维数组a[3][3]的各个元素:\n");for (i = 0; i < 3; i++){ for (j = 0; j < 3; j++){ scanf("%d", &a[i][j]);} } printf("输入一个整数:\n");scanf("%d", &num);for ...

include <iostream.h> void main(){ int a[2][3];//判断循环次数_即数组元素:2+1+3+1=7 int b;for(int n=0;n<=2;n++){//从键盘接受数据对此二维数组初始化 for(int m=0;m<=3;m++){ cout<<"请输入a"<<n<<m<<endl;cin>>a[n][m];//请输入整型数值,否则会失去精度或者...

a1[j][i]=a1[i][j];a1[i][j]=t;} } main(){ int a[3][3]={{1,2,3},{4,5,6},{7,8,9}};int i,j;f(a);printf("转置后的数组为\n");for(i=0;i<3;i++){ for(j=0;j<3;j++)printf("%d ",a[i][j]);printf("\n");} system("PAUSE");return 0;} ...

include <stdio.h>int main(){ int a[3][4]; int l[3]={0},c[4]={0},s=0; int i,j; for(i = 0; i < 3;i++) { for(j = 0; j < 4; j ++) { scanf("%d",&a[i][j]); s+=a[i][j]; l[i]+=a[i][j]; c[j]+=a...

max = arr[i][j];} } } printf("最大值为：%d\n", max);return 0;} 在上面的代码中，我们首先定义了一个3行4列的二维整型数组 arr，并将其初始化为特定的值。接着，我们定义一个变量 max，用于存储数组中的最大值，我们假设数组的第一个元素为最大值。然后，我们使用两个嵌套的循环来...