样例输入:
8
2 8 3 5 1 7 3 6
4
-4 6 10 0
样例输出:
Case 1: 1 8 4.375
Case 2: -4 10 3.000
#include <stdio.h>
#define INF 1000000000
int main()
{
int x, n = 0, ***kase = 0;***
while(scanf("%d", &n) == 1 ***&& n***){
***int s = 0;***
int min = INF;
int max = INF;
for(int i = 0; i < n; i++){
scanf("%d", &x);
s += x;
if(x < min) min = x;
if(x > max) max = x;
}
***if(kase) printf("\n");***
printf("Case %d; %d %d %.3f\n", ***++kase***, min, max, (double)s/n);
}
return 0;
}
- 用kase变量输出Case n:
- 判断是否结束用 && n