#include <stdio.h>
void main()
{
void print(int *a);
void reverse(int *arr,int n,int m);
int n,i,m=4;
int a[10];
printf("input N:");
scanf("%d",&n);
printf("please input %d numbers:\n",n);
for(i=0;i<n;i++)
{
scanf("%d",(a+i));
}
print(a);
reverse(a,n,m);
print(a);
}
void print(int *a)
{
int i;
for(i=0;i<10;i++)
{
printf("%d ",*(a+i));
}
}
void reverse(int *arr,int n,int m)
{
void inverse(int *arr,int to,int from);
inverse(arr,0,n-1);
inverse(arr,0,m-1);
inverse(arr,m,n-1);
}
void inverse(int *arr,int to,int from)
{
int i,t;
for(i=0;i<=(from-to)/2;i++)
{
t = *(arr+to+i);
*(arr+to+i) = *(arr+from-i);
*(arr+from-i) = t;
}
}
总结:inverse函数中,to加到from的值,是(from-to)/2
结果:input N:10
please input 10 numbers:
0 1 2 3 4 5 6 7 8 9
6 7 8 9 0 1 2 3 4 5