【题意分析】1024 Palindromic Number (25 分)

立志用最少的代码做最高效的表达

PAT甲级最优题解——>传送门

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

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. For example, if we start from 67, we can obtain a palindromic number in 2 steps: 67 + 76 = 143, and 143 + 341 = 484.

Given any positive integer N, you are supposed to find its paired palindromic number and the number of steps taken to find it.

Input Specification:
Each input file contains one test case. Each case consists of two positive numbers N and K, where N (≤10^10) is the initial numer and K (≤100) is the maximum number of steps. The numbers are separated by a space.

Output Specification:
For each test case, output two numbers, one in each line. The first number is the paired palindromic number of N, and the second number is the number of steps taken to find the palindromic number. If the palindromic number is not found after K steps, just output the number obtained at the Kth step and K instead.

Sample Input 1:
67 3
Sample Output 1:
484
2

Sample Input 2:
69 3
Sample Output 2:
1353
3

题意：给定数N，次数K，将N翻转后与原数相加，判断是否为回文数，如果是则直接输出，如果不是，则算作一次操作，操作次数上限为K。若K次后仍不是回文数，则直接输出该数，结束循环。

分析：利用了字符串函数翻转机制缩减了代码量。代码逻辑很简单，读者可自行体会。

#include<bits/stdc++.h>
using namespace std;bool is_palindormic(string s) {     //判断是否是回文数 string s1 =s;reverse(s.begin(), s.end());if(s1 == s) return true;else return false;
}string ope (string s) {                //执行相加操作 string s1 = s, add_s;reverse(s.begin(), s.end());int carry = 0, x;for(int i = 0; i < s.size(); i++) {x = (s[i]-'0' + s1[i]-'0' + carry) % 10;carry = (s[i]-'0' + s1[i]-'0' + carry) / 10;add_s += x+'0';}if(carry > 0) add_s += carry+'0';reverse(add_s.begin(), add_s.end());return add_s;
}int main() {string s; int n, n1;cin >> s >> n;n1 = n;while(n1) if(is_palindormic(s)) {    //如果是回文数，则输出，结束循环 cout << s << '\n' << n-n1 << '\n'; goto loop;} else {               //如果不是回文数， n1--;s = ope(s);}cout << s << '\n' << n << '\n'; loop : ;return 0;
}

耗时：

【题意分析】1024 Palindromic Number (25 分)_38行代码AC相关推荐

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【题意分析】1024 Palindromic Number (25 分)_38行代码AC

立志用最少的代码做最高效的表达

耗时：

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