计算机考试-C 回文计算—东方仙盟

Use VC++2010 to open the solution in the candidate folder blank1. The project contains one source file blank1.c. The function fun() judges whether the string pointed to by parameter s is a palindrome . If it is a palindrome, the function returns 1. If not, it returns 0. A palindrome reads exactly the same forward and backward, case‑insensitive. For example: "LEVEL" and "Level" are palindromes, while "LEVLEV" is not.

Complete the program by filling in the blanks. Do not add or delete lines, do not change program structure.

#include <stdio.h>
#include <string.h>
#include <ctype.h>

int fun(char * s)
{
    char * lp, * rp;
    /**********found**********/
    lp = 【1】 ;
    rp = s + strlen(s) - 1;

    while( (toupper(*lp)==toupper(*rp)) && (lp < rp) )
    {
        /**********found**********/
        lp ++; rp 【2】 ;
    }

    /**********found**********/
    if (lp < rp) 【3】 ;
    else return 1;
}

main()
{
    char s[81];
    printf("Enter a string: ");
    scanf("%s",s);
    if(fun(s))
        printf("\n\"%s\" is a Palindrome.\n\n",s);
    else
        printf("\n\"%s\" isn't a Palindrome.\n\n",s);
}

Answers

【1】s 【2】-- 【3】return 0

---

Explanation

Return value rule

• return 1: It is a palindrome.
• return 0: It is NOT a palindrome.

The program uses two pointers:

• lp: left pointer, starts at the first character of string.
• rp: right pointer, starts at the last character of string.

while( (toupper(*lp)==toupper(*rp)) && (lp < rp) ) Loop runs only when two conditions are both true:

1. The uppercase of left‑side character equals uppercase of right‑side character.
2. Left pointer is still before right pointer (lp < rp).

Inside loop: move left pointer forward, move right pointer backward toward the middle.

When the while loop stops:

1. If lp < rp is still true → we met different characters. Not palindrome → return 0.
2. If pointers meet or cross each other → all matched. Is palindrome → return 1.

---

Test case 1: Even‑length palindrome ABCCBA（偶数长度）

Characters: A B C C B A

• lp points to A, rp points to A → equal, lp++, rp--
• lp points to B, rp points to B → equal, lp++, rp--
• lp points to C, rp points to C → equal, lp++, rp-- Now lp is at index 3, rp is at index 2. lp < rp → 3 < 2 → false. While loop exits. Skip if(lp<rp), execute else return 1. Result: it is a palindrome.

Test case 2: Odd‑length palindrome ABCBA（奇数长度）

Characters: A B C B A

• lp=0(A), rp=4(A): equal → lp++, rp--
• lp=1(B), rp=3(B): equal → lp++, rp-- Now lp and rp both point to middle character C(index 2). Enter while‑loop: characters equal, lp<rp (2<2) is false. Loop condition fails. Loop stops directly. Execute else return 1. Result: it is a palindrome.

One non‑palindrome example ABCDE

First compare A vs E: not equal. The while loop stops immediately. lp < rp holds true. Run return 0. Output: not a palindrome.

Why return 0? 0 stands for logical false in C language, meaning the proposition "this string is palindrome" is false. 1 stands for logical true, meaning the proposition holds true.

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