Question

用C语言编写一个程序读取输入，直到#，并报告ei出现的次数

我所写的程序如下，让我不满意的地方是用到了goto，我想要不用goto语句完成这个任务。
#include <stdio.h>
int main(void)
{
char ch,cur;
int n=0;

while((ch=getchar())!='#')
{
fir:if(ch=='e')
{
cur=getchar();
if(cur=='i')
n++;
else
{
ch=cur;
goto fir;
}
}

}
printf("%d",n);
return 0;
}

Answer 1

#include<stdio.h>
int main(void)
{
	char ch;
	int e=0,n=0;
	while( (ch=getchar())!='#' ){
		if( ch=='e' )
			e=1;
		else{
			if( ch=='i' && e==1 )n++;
			e=0;
		}
	}
	printf( "%d\n", n );
	return 0;
}

Answer 2

#include <stdio.h>
int main(void)
{
char ch;
char last_ch = 0;
int count=0;

printf("Enter some texts(# to quit):\n");
while ((ch = getchar()) != '#')
{
if ((ch == 'i') && (last_ch == 'e'))
count++;
last_ch = ch; //出现ei时，计数+1
}
printf("\"ei\" appeared %d times.\n", count);

return 0;
}

Answer 3

#include<stdio.h>
int main(void)
{
char ch;
int e=0,n=0;
while( (ch=getchar())!='#' ){
if( ch=='e' )
e=1;
else{
if( ch=='i' && e==1 )n++;
e=0;
}
}
printf( "%d\n", n );
return 0;
}

Answer 4

#include<stdio.h>
int main()
{
char ch;
int n=0;
while((ch=getchar())!='#')
{
while('e'==ch)
{
if('i'==(ch=getchar()))
{
n++;
}
}
}
printf("%d\n",n);
return 0;
}