View Code of Problem 9

#include<stdio.h>
#include<string.h>
void main()
{
	char a[50];
	int n,sum=0,count[5]={0},k=1;
	while(scanf("%d",&n)!=EOF)
	{
		for(int i=0;i<=n;i++)
		{
			gets(a);
			if(strcmp("Pleasant goat",a)==0)
				count[0]++;
			else if(strcmp("Pretty goat",a)==0)
				count[1]++;
			else if(strcmp("Athletic goat",a)==0)
				count[2]++;
			else if(strcmp("Lazy goat",a)==0)
				count[3]++;
			else if(strcmp("Slow goat",a)==0)
				count[4]++;
		}
		sum=5*count[0]+8*count[1]+10*count[2]+15*count[3]+20*count[4];
		if(n>=10&&count[2]>0)
			sum+=50;
		printf("Case #%d:%d\n",k++,sum);
		for(int j=0;j<5;j++)//每一次计算结束后将count数组元素都置为零，否则技计数数组会累加，导致出错
			count[j]=0;
	}
}