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 ai as ak⋯a1a0 with 0≤ai<10 for all i and ak>0. Then N is palindromic if and only if ai=ak−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 ak⋯a1a0 的形式,其中对所有 i 有 0≤ai<10 且 ak>0。N 被称为一个回文数,当且仅当对所有 i 有 ai=ak−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
- 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;
- 以字符串形式读入待检查数字;
- 检查合格则输出,否则执行字符串相加,并计数;
- 循环到合格或达到十次后输出结果;
- 返回零值。
代码
[code]#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; }
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