#include "stdio.h"
#define N 5 //5可以修改成你要是数字!
main()
{int i,a[N-1];
for(i=0;i
scanf("%d",&a[i]);
}
for(i=N-1;i>=0;i--)
printf("a[%d]=%d\n",i,a[i]); //输出数组
}
看了一下 上面的6个回答是不是跟要求不符阿
上面的回答只是“输入一组整数,将其顺序颠倒输出”
而要求是“将其顺序颠倒后输出”注意“后”
我写一个参考一下吧:
#include
int main(void)
{
int Num;
int i ;
int j ;
int a_To_b;
int a[100];
printf("How many integers do you want to erter (Num < 100):\n");
scanf("%d",&Num);
if(Num >=100)
{
printf("It is a bad number !\n");
return 1;
}
printf("\n");
printf("Please erter :\n");
for(i=0;i
printf("Enter the value of a[%d] ",i);
scanf("%d",&a[i]);
}
printf("\n");
printf("Now begin to reverse the a[%d]: \n",Num);
for(i=0,j=Num-1; i<(Num/2); i++,j--)
{
a_To_b = a[i];
a[i] = a[j];
a[j] = a_To_b;
}
printf("reversal is finished:\n");
printf("\n");
printf("Now output the a[%d]:\n",Num);
for(i=0;i
printf("a[%d] = %d\n",i,a[i]);
}
return 0;
}
main()
{
int a[10],i;
printf("Input:\n");
for(i=0;i<10;i++)
{
scanf("%d",&a[i]);
}
printf("\nOutput:\n");
for(i=9;i>=0;i--)
{
printf("%d\n",a[i]);
}
}
这种题目有两种做法,我告诉你一个简单的:
main()
{
int a[20],i;
for(i=0;i<20;i++) /*输入数字,假设是20个*/
{
scanf("%d",&a[i]);
}
for(i=19;i>=0;i--) /*输出数字*/
{
printf("&d",a[i]);
}
}
main()
{
int a[20],i;
for(i=0;i<20;i++) //输入数字,假设是20个
{
scanf("%d",&a[i]);
}
for(i=19;i>=0;i--) //输出数字
{
printf("&d",a[i]);
}
}