用c语言编写一个程序，输入一个字符串，统计出某指定字符在字符串中出现的次数？

Question

如题所述

Answer 1

#include "stdio.h"
void main()
{
int i,a=0,b=0,c=0,d=0,n=0;/*分别用于统计英文字符、空格、数字、其它字符、所有字符的统计*/
printf("Please input a string!\n");
while(getchar()!='\n')
n++;
i = getchar();
if(65<=i<=90||97<=i<=122)
a++;
else if(i=0)
b++;
else if(48<=i<=57)
c++;
else
d++;
printf("所输入的字符串中英文字符有%d个，空格有%d个，数字有%d个，其它字符有%d个，总共%d个!\n",a,b,c,d,n);
return 0;
}

Answer 2

#include <stdio.h>
void main()
{
char s,c[10];
int i,n=0;
printf("输入一字符串:\n");
gets(c);
printf("输入指定字符s:\n");
s=getchar();
for(i=0;i<=10;i++)
if(c[i]==s)
n++;
printf("%d\n",n);
}

参考资料：黄保和

Answer 3

#include <stdio.h>
#include <string.h>
void main()
{char c,s[80]; int i,k=0;
gets(s); /*输入字符串*/
printf("Please input a character:"); c=getchar();
for(i=0; s[i]!='\0'; i++)
{if(s[i]==c)
k++;
}
printf("%c is found %d times",c,k);
}本回答被提问者和网友采纳

Answer 4

#include <stdio.h>
#include <string.h>
void main(void)
{
char str[80],s,iCount=0;
int i;
printf("please input the string.\n");
gets(str);
printf("please input the character that you want to count.\n");
s=getchar();
for(i=0;str[i]!='\0';i++)
{
if(str[i]==s)
{
iCount++;
}
}
printf("icount=%d\n",iCount);

}