Question

C语言问题，为什么会显示Runtime Error

AfterShawn sees the following picture, he decides to give up his career in IT andturn to sell fruits. Since Shawn is a lazy guy, he doesn’t like to do any extrawork. When he starts selling fruits, he finds that he always needs to takechanges for customers. He wants you to write a program to give the least numberof changes. Shawn is also a strange guy, because he takes changes by 6,5,3,1.

Input
Thefirst line of input is an integer T which indicates the sum of test case. Eachof the following T lines contains an integer n(1≤n≤105), which isthe total money Shawn need to give to the customer.

Output
Foreach test case, output "Case i: Result" in one line whereiis the case number andResultis the least number of changes Shawnneed to give customer.

SampleInput
3
4
10
19

Sampleoutput
Case 1: 2
Case 2: 2
Case 3: 4

下面是我的：

#include <stdio.h>

int main()
{
int i,j,a[10],s;
scanf("%d",&j);
for(i=0;i<j;i++)
{
scanf("%d",&a[i]);
}

for(i=0;i<j;i++)
{
if(a[i]%5==0)
s=a[i]/5;
else
s=a[i]/6+(a[i]%6)/5+((a[i]%6)%5)/3+((a[i]%6)%5)%3;

printf("Case %d: %d\n",i+1,s);
}

return 0;
}

Answer 1

应该是没有什么问题啊

追问

对啊，没问题啊，但是提交到ACM网址的那里，显示的是

搞的我很恼火。。三小时就只琢磨这一道题了

追答

ACM的题，嗯，我详细看了下，
如果你的j大于10时，程序的确是runtime error的，这个问题只能用动态内存分配来做：
#include "stdio.h"
#include "stdlib.h"
int main()
{
int i,j,s;
int *a=NULL;
scanf("%d",&j);
a=(int *)malloc(sizeof(int)*j);
for(i=0;i<j;i++)
{
scanf("%d",a+i);
}

for(i=0;i<j;i++)
{
if(a[i]%5==0)
s=a[i]/5;
else
s=a[i]/6+(a[i]%6)/5+((a[i]%6)%5)/3+((a[i]%6)%5)%3;

printf("Case %d: %d\n",i+1,s);
}
free(a);
a=NULL;
return 0;

}
这样才是最对的做法，你可以用这个去提交

追问

额额额额额。。不会是这个样子吧。。。。坑死人啊，，那是考试，不能提交了。。如果我改成a[100]不就好了么

追答

当然不行了，
改100也是不行的，
因为你不能确定到底有多少顾客，所以只能动态根据你输入的顾客数来确定，
这样，没有空间浪费，
估计他考虑其中的一个方面就在这里。