C语言“回文”程序代码

首先我对你的 "并且当输入的字符串第一个字符为＃时，输入为空时，不输出。" 这句话比较费解, 不明白你的意思!

你把这说清楚了我再补充回答你~

我写了个参考代码给你参考, 首先是输入你要谈差判断的字符串的个数, 然后再依次输入所有的字符串, 最后判断输入的所有字符串是否是"回文"! 因为不理解你那句话, 所以暂时没做什么空和什么"#"处理.

详细c代码:

#include <stdio.h>

#include <string.h>

#define STR_LEN 128

#define STR_NUM 64

int main()

{

int i = 0, j = 0, n = 0

int len = 0

char *start = NULL

char *end = NULL

char str[STR_NUM][STR_LEN] = {0}

printf("Please input the number of string: \n")

scanf("%d", &n)

printf("Please input all the string: \n")

for (i = 0i <ni++)

{

scanf("%s", str[i])

}

for (j = 0j <nj++)

{

start = str[j]

len = strlen(str[j])

end = str[j] + len - 1

while (start - end <= 0)

{

if (*start++ != *end--)

{

break

}

}

if (start >end)

{

printf("yes\n")

}

else

{

printf("no\n")

}

}

return 0

}

例子,运行后:

Please input the number of string:

4

Please input all the string:

aba

112ds

madam

xyzyx

yes

no

yes

yes

Press any key to continue

补充回答:

大森兄概明白你的意思,我会此侍袭把例子贴上, 若不符合你的要求我再帮修改!

详细代码如下:

#include <stdio.h>

#include <string.h>

#define STR_LEN 128

#define STR_NUM 64

int main()

{

int i = 0

int len = 0

int str_count = 0

char *start = NULL

char *end = NULL

char str[STR_NUM][STR_LEN] = {0}

printf("Please input all the string: \n")

while (1)

{

scanf("%s", str[str_count])

if (str[str_count][0] == '#')

{

break

}

else

{

str_count++

continue

}

}

for (i = 0i <str_counti++)

{

start = str[i]

len = strlen(str[i])

end = str[i] + len - 1

while (start - end <= 0)

{

if (*start++ != *end--)

{

break

}

}

if (start >end)

{

printf("yes\n")

}

else

{

printf("no\n")

}

}

return 0

}

运行实例1:

Please input all the string:

xyzyx

adghf

#

yes

no

Press any key to continue

运行实例2:

Please input all the string:

1232ss

sakljfkla

333dafs

aba

ee3

xyzyx

dfj222

madam

111$111

slsl33

#

no

no

no

yes

no

yes

no

yes

yes

no

Press any key to continue

有分多多给啊，我给你费了不少时间呢。程序完全没有问题，你所说的功能都笑肢给你做到了哦！槐升袭

#include<stdio.h>

#include<string.h>

int main()

{

char buf[1000]={0}

printf("Please input :\n")

int i,j,k=0

for(i = 0i<1000i++){

buf[i] = getchar()

if(buf[i] == '\n' )

break

}

printf("%s\n",buf)

for(i = 0 i<1000 i++ ){

j = buf[i]

if( ( j>=48 &&j<=57 )|| (j>=65 &&j<=90 ))

buf[k++] = j

else if((j>=97 &&j<=122) )

buf[k++] = j-32

else if(j == '\n' ){

buf[k] = '\0'

break

}

}

printf("%s\n",buf)

j = strlen(buf)

for(i = 0 i <strlen(buf)/2 i++,j--){

if( buf[i] != buf[j-1] )

break

}

if(i != strlen(buf)/2 )

printf("No i=%d\n",i)

else

printf("Yes\铅兄n")

}

#include <stdio.h>

int palindrome(char *string)

void main()

{int i

char str[100],*string=str

gets(string)

i=palindrome(string)

printf("i=%d\n",i)

getch()

}

int palindrome(char *string)

{int n=0,i,flag=1

char *p=string

while(*p!='凯拍\物猛0'盯蚂羡)

{n++p++}

for(i=0i<=n/2i++)

{if(*(string+i)==*(string+n-i-1))

else return 0

}

return 1}

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