PAT-A1136/B1079 A Delayed Palindrome/延迟的回文数 题目内容及题解

Consider a positive integer N written in standard notation with k+1 digits a​i​​ as a​k​​⋯a​1​​a​0​​ with 0≤a​i​​<10 for all i and a​k​​>0. Then N is palindromic if and only if a​i​​=a​k−i​​ for all i. Zero is written 0 and is also palindromic by definition.

Non-palindromic numbers can be paired with palindromic ones via a series of operations. First, the non-palindromic number is reversed and the result is added to the original number. If the result is not a palindromic number, this is repeated until it gives a palindromic number. Such number is called a delayed palindrome. (Quoted from https://en.wikipedia.org/wiki/Palindromic_number )

Given any positive integer, you are supposed to find its paired palindromic number.

给定一个 k+1 位的正整数 N，写成 a​k​​⋯a​1​​a​0​​ 的形式，其中对所有 i 有 0≤a​i​​<10 且 a​k​​>0。N 被称为一个回文数，当且仅当对所有 i 有 a​i​​=a​k−i​​。零也被定义为一个回文数。

非回文数也可以通过一系列操作变出回文数。首先将该数字逆转，再将逆转数与该数相加，如果和还不是一个回文数，就重复这个逆转再相加的操作，直到一个回文数出现。如果一个非回文数可以变出回文数，就称这个数为延迟的回文数。（定义翻译自 https://en.wikipedia.org/wiki/Palindromic_number ）

给定任意一个正整数，本题要求你找到其变出的那个回文数。

Input Specification:

Each input file contains one test case which gives a positive integer no more than 1000 digits.

输入在一行中给出一个不超过1000位的正整数。

Output Specification:

For each test case, print line by line the process of finding the palindromic number. The format of each line is the following:

对给定的整数，一行一行输出其变出回文数的过程。每行格式如下：

A + B = C

where A is the original number, B is the reversed A, and C is their sum. A starts being the input number, and this process ends until C becomes a palindromic number -- in this case we print in the last line C is a palindromic number.; or if a palindromic number cannot be found in 10 iterations, print Not found in 10 iterations. instead.

其中 A 是原始的数字，B 是 A 的逆转数，C 是它们的和。A 从输入的整数开始。重复操作直到 C 在 10 步以内变成回文数，这时在一行中输出 C is a palindromic number.；或者如果 10 步都没能得到回文数，最后就在一行中输出 Not found in 10 iterations.。

Sample Input 1:

97152

Sample Output 1:

97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.

Sample Input 2:

196

Sample Output 2:

196 + 691 = 887
887 + 788 = 1675
1675 + 5761 = 7436
7436 + 6347 = 13783
13783 + 38731 = 52514
52514 + 41525 = 94039
94039 + 93049 = 187088
187088 + 880781 = 1067869
1067869 + 9687601 = 10755470
10755470 + 07455701 = 18211171
Not found in 10 iterations.

解题思路

1. 分别建立检查字符串是否回文、翻转字符串和字符串内容相加（包含输出中间过程）的函数；
2. 以字符串形式读入待检查数字；
3. 检查合格则输出，否则执行字符串相加，并计数；
4. 循环到合格或达到十次后输出结果；
5. 返回零值。

代码

#include<stdio.h>
#include<string.h>
#define maxn 1050
char ans[maxn],re[maxn],sum[maxn];

int check (char a[]){
    int left=0,right=strlen(a)-1;
    while(left<right){
        if(a[left]!=a[right]){
            return 0;
        }
        left++;
        right--;
    }
    return 1;
} //1对称

void Reverse(char a[],char b[]){
    int i=0,len=strlen(a)-1;
    while(len>=0){
        b[i++]=a[len];
        len--;
    }
    b[i]=0;
    return;
}

void Add(){
    int i=0,c=0,num;
    Reverse(ans,re);
    while(ans[i]!=0){
        num=re[i]+ans[i]-2*'0'+c;
        c=num/10;
        sum[i++]=num%10+'0';
    }
    if(c){
        sum[i++]='1';
    }
    sum[i]=0;
    printf("%s + %s = ",ans,re);
    Reverse(sum,ans);
    printf("%s\n",ans);
}

int main(){
    int N=0;
    scanf("%s",ans);
    while(1){
        if(check(ans)){
            printf("%s is a palindromic number.\n",ans);
            return 0;
        }
        Add();
        N++;
        if(N>=10){
            break;
        }
    }
    printf("Not found in 10 iterations.\n");
    return 0;
}

运行结果