PAT甲级1024 ASCII码与整数转换

最新推荐文章于 2023-12-28 11:32:39 发布

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最新推荐文章于 2023-12-28 11:32:39 发布

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分类专栏：刷子刷题文章标签： PAT ASCII码

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本文介绍了一种算法，用于寻找任意正整数N的配对回文数及其所需步骤。通过反转并累加非回文数直到形成回文数，文章详细解释了这一过程，并提供了完整的C++代码实现。

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题目

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

题解

#include<iostream>
#include<algorithm>
using namespace std;
string s;

void add(string t){
  int len=s.length();
  int flag=0;
  for(int i=len-1;i>=0;i--){//这边循环我把末位作为最低位
    s[i]=(s[i]-'0')+(t[i]-'0')+flag+'0';//得到字符（这边柳神的博客简化了，我扩充回来）
    flag=0;
    if(s[i]>'9'){
      s[i]=s[i]-10;//ASCII码内部转换为int，与10相减
      flag=1;
    }
  }
  if(flag) s='1'+s;//最后一位有进位，字符串前面+‘1’
  
}
int main(){
  int cnt,i;
  cin>>s>>cnt;
  
for(i=0;i<cnt;i++){//这边的i不能重新定义！！否则输出i为随机数
  string t=s;
  reverse(t.begin(),t.end());
  if(s==t||i==cnt) break;
  add(t);

}
  cout<<s<<endl<<i;
  return 0;
}

这题的矛盾点主要在，我以为s[i]超出‘9’后，应该减17得到个位数。经测试。‘7’+‘6’输出了‘=’ ，而不是D。

‘=’的ASCII码为13，变为3。因此还是减去10。