将一个3*3的矩阵转置,用一个函数实现。在主函数中输入以下矩阵元素:{2,4,6,8,10,12,14,16,18}。将数组名作为函数参数。函数调用后在主函数中输出已转置的矩阵
#include<stdio.h>
#include<iostream>
using namespace std;
#define N 3
void f(int a[N][N],int n) { int i,j,k;
for ( i=0;i<N;i++ ) for ( j=0;j<i;j++ ) {k=a[i][j];a[i][j]=a[j][i];a[j][i]=k;}
}
void main() { int i,j,a[N][N];//={2,4,6,8,10,12,14,16,18};
for ( i=0;i<N;i++ ) for ( j=0;j<N;j++ ) cin>>a[i][j]; //scanf("%d",&a[i][j]); printf("\n");
cout <<endl;
for ( i=0;i<N;i++,cout<<endl /*printf("\n")*/ )
for ( j=0;j<N;j++ ) cout<<a[i][j]<<" "; //printf("%2d ",a[i][j]);
f(a,N);
cout<<"**************"<<endl; //printf("**************\n");
for ( i=0;i<N;i++,cout<<endl /*printf("\n")*/ )
for ( j=0;j<N;j++ ) cout<<a[i][j]<<" "; //printf("%2d ",a[i][j]);
}
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追答ok
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