1136. A Delayed Palindrome (20)
来源:互联网 发布:血液问题人数调查数据 编辑:程序博客网 时间:2024/06/06 11:47
说明:
1、测试数据稍弱;如果输入10000,应该输出10000 + 1 = 10001,而不是+ 00001,但本测试用例无此情况,故可以略去
2、实际考察大数相加
注意:
1、如果输入A本身是回文数,则不进行迭代求和,应该直接输出结果
#include <bits/stdc++.h>using namespace std;string add(const string &A,const string &B){ int i=A.size()-1,sur=0; string C; while(i>=0) { int sum=A[i]-'0'+B[i]-'0'; C+=(sur+sum)%10+'0'; sur=(sur+sum)/10; --i; } if(sur>0)C+='0'+sur; reverse(C.begin(),C.end()); return C;}bool ispalindromic(const string &C){ int len=C.size(),i; for(i=0;i<len/2;++i) if(C[i]!=C[len-1-i]) return false; return true;}int main(){ string A,B,C; int cnt=10; cin>>A; if(ispalindromic(A)) { cout<<A<<" is a palindromic number."<<endl; return 0; } while(cnt--) { B=A; reverse(B.begin(),B.end()); C=add(A,B); cout<<A<<" + "<<B<<" = "<<C<<endl; if(ispalindromic(C)) { cout<<C<<" is a palindromic number."<<endl; return 0; } A=C; } cout<<"Not found in 10 iterations."<<endl; return 0;}
阅读全文
0 0
- 1136. A Delayed Palindrome (20)
- pat-1136. A Delayed Palindrome (20) 模拟
- 1136. A Delayed Palindrome (20) 大数模拟
- PAT甲级1136. A Delayed Palindrome (20)
- Valid a Palindrome
- A. Mike and palindrome
- A. Mike and palindrome
- A.Quasi-palindrome
- Delayed Project
- Delayed Messaging
- Delayed ACK
- Codeforces 863A Quasi-palindrome
- codeforces 863A Quasi-palindrome
- Check whether a string is a palindrome
- decide weather a string is a palindrome
- CareerCup Facebook Judge Whether a K-palindrome
- Check if a linked list is palindrome
- Minimum insertions to form a palindrome
- web.xml 中的 servlet 和 servlet-mapping 标签
- httpClient入门到精通-------连接池的关闭
- kinect v2_bodyindex
- leetcode 525. Contiguous Array 统计1和0数量相等的最长子数组
- C++之cmath常用库函数一览
- 1136. A Delayed Palindrome (20)
- redis学习--下载、安装
- Tomcat中 HttpServletRequst的获取网页的几种方法比较
- 简单说一说Java的内存泄漏
- LINQ、反射和特性
- 迷宫问题
- 选择排序和插入排序
- machine learning实践学习二:K-Nearest Neighbors
- 场次降雨数据划分