C语言，设计一个函数，判断一个字符串是否回文，即顺读和倒读的结果都一样。若是，返回1；若否，返回0.

Question

如题所述

Answer 1

#include "stdlib.h"
#include "stdio.h"
int f(char a[],int n);
int main()
{
int n,i;
scanf("%d",&n);
if(n<1)
return 1;
char *a = (char*) malloc(n*sizeof(char));
char c;
do
{
scanf("%c",&c);
}while(c=='\n');
a[0] = c;
for (i=1;i<n;i++)
scanf("%c",a+i);
printf("%d",f(a,n));
free(a);
system("pause");
return 0;
}
int f(char a[],int n)
{
int i;
for (i=0;i<n/2;i++)
{
if(a[i] != a[n-i-1])
return 0;
}
return 1;
}

Answer 2

int exam(char *str)
{
char *p=str+strlen(str)-1;
while(str<p)
{
if(*str!=*p)
{
return 0;
}
str++;
p--;
}
return 1;
}

Answer 3

#include "stdafx.h"
#include<iostream>
#include<iomanip>
using namespace std;
bool compare(const char *a)
{
char b[100];
strcpy(b,a);
strrev(b);
if(strcmp(a,b)==0)
return 1;
else
return 0;
}
int _tmain(int argc, _TCHAR* argv[])
{
char ch[100];
gets(ch);
cout<<compare(ch)<<endl;
return 0;
}

Answer 4

int exam ()
{
int c,i;
c=strlen(s);
for(i=0;i<=c;i++)
{
c--;
if(str[i]!=str[c])
return 0;
}
return 1;
}

Answer 5

#include <stdio.h>
#include <string.h>
void main()
{
int i,k;
long n,m;
int digit[10];
puts("input a position integer:");
scanf("%d",&n);
m=n;k=0;
do
{
digit[k++]=m%10;
m/=10;
}while(m!=0);
k--;
for(i=0;i<k;i++,k--)
if(digit[i]!=digit[k]) break;
if(i<k) printf("%ld is not a palindrome.",n);
else printf("%ld is a palindrome.",n);
}本回答被提问者采纳