编一函数将一维整型数组a[5]={1,3,5,7,9}的元素倒置存放。(要求使用指 ...答:int i,a[5]={1,3,5,7,9},*p0=a,*p1=&a[4];while(p0<p1)p0^=*p1,*p1^=*p0,*p0++^=*p1--;//打印倒置后的数组 for(i=0;i<5;printf("%d ",a[i]),i++);return 0;}
用C++编写一个把有N个元素的数组a中的元素逆置保存到数组b的函数,并设 ...答:include "stdio.h"void fun(int a[],int b[],int n){ int i;for(i=0;i<n;i++)b[i]=a[n-1-i];} void main(){ int a[5]={1,2,3,4,5};int b[5]={0};int i;fun(a,b,5);for(i=0;i<5;i++)printf("%d ",b[i]);printf("\n");} 结果:
用C语言设计 数组倒置。子函数原型为:void fun(int a[],int n),其功...答:include<stdio.h> void fun(int a[],int n){ int t;for(int i=0;i<n/2;i++){ t=a[i];a[i]=a[n-i-1];a[n-i-1]=t;} } void main(){ int a[10]={0,1,2,3,4,5,6,7,8,9};fun(a,10);for(int i=0;i<10;i++)printf("%d ",a[i]);printf("\n");}...
...reverse(int a[ ],int n) 颠倒数组a的前n个元素的顺序答:void reverse(int *a,int n){ int i,t;for(i=0;i<n/2;i++){ t=*(a+i);(a+i)=*(a+n-i-1);(a+n-i-1)=t;} } int main(){ int i,a[]={0,1,2,3,4,5,6,7,8,9};reverse(a,10);for(i=0;i<10;i++)cout<<a[i]<<' ';cout<<endl;return 0;} ...