Codeforces Round #342 (Div. 2) D. Finals in arithmetic(想法题/构造题)
Description
Vitya is studying in the third grade. During the last math lesson all the pupils wrote on arithmetic quiz. Vitya is a clever boy, so he managed to finish all the tasks pretty fast and Oksana Fillipovna gave him a new one, that is much harder.
Let's denote a flip operation of an integer as follows: number is considered in decimal notation and then reverted. If there are any leading zeroes afterwards, they are thrown away. For example, if we flip 123 the result is the integer 321, but flipping 130 we obtain 31, and by flipping 31 we come to 13.
Oksana Fillipovna picked some number a without leading zeroes, and flipped it to get number ar. Then she summed a and ar, and told Vitya the resulting value n. His goal is to find any valid a.
As Oksana Fillipovna picked some small integers as a and ar, Vitya managed to find the answer pretty fast and became interested in finding some general algorithm to deal with this problem. Now, he wants you to write the program that for given n finds any a without leading zeroes, such that a + ar = n or determine that such a doesn't exist.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 10100 000).
Output
If there is no such positive integer a without leading zeroes that a + ar = n then print 0. Otherwise, print any valid a. If there are many possible answers, you are allowed to pick any.
Sample Input
4 11
5 33
Sample Output
2 10 0 21
Note
In the first sample 4 = 2 + 2, a = 2 is the only possibility.
In the second sample 11 = 10 + 1, a = 10 — the only valid solution. Note, that a = 01 is incorrect, because a can't have leading zeroes.
It's easy to check that there is no suitable a in the third sample.
In the fourth sample 33 = 30 + 3 = 12 + 21, so there are three possibilities for a: a = 30, a = 12, a = 21. Any of these is considered to be correct answer.
思路
题意:给出数字n,问是否存在一个数x,使得 x + flip(x) = n (其中 flip(x)为x的反转,反转后忽略前导0)
题解:对于前后两个对称位置,如果相等,则ans[i]=(num[i]+1)/2,ans[n-i-1]=num[i]/2,若不相等,考虑两种情况,一种是来自低位的进位,一种是来自高位的退位
- 当num[i] == num[n - i - 1] + 1 || num[i] == num[n - i - 1] + 11,说明第 i 位有来自低位的进位,因此将其还原即可,亦即使得num[i]--,num[i + 1] += 10;
- 当num[i] == num[n - i - 1] + 10,说明第 i 位有来自高位的退位,因此使得第 n - i - 1 位也有来自高位的退位,,亦即使得num[n - i - 2]--,num[n - i - 1] += 10;
另外,对于n为 “1”开头的数值时,需要特判,因为这位1是由x 和 flip(x )相加而得的进位,至于大于 “1”的数字开头的数值不需要进位是因为 9 + 9 + 1 = 19,最多只能进1。
#include<bits/stdc++.h> using namespace std; const int maxn = 1000005; char s[maxn],res[maxn]; int num[maxn]; bool check(int n) { for (int i = 0;i < n / 2;) { if (num[i] == num[n - i - 1]) i++; else if ((num[i] == num[n - i - 1] + 1) || (num[i] == num[n - i - 1] + 11)) //来自低位的进位 { num[i]--; num[i + 1] += 10; } else if (num[i] == num[n - i - 1] + 10) //来自高位的退位 { num[n - i - 2]--; num[n - i - 1] += 10; } else return false; } if (n % 2 == 1) { if ((num[n/2]%2 == 1) || (num[n/2] > 18) || (num[n/2] < 0)) return false; else res[n/2] = num[n/2]/2 + '0'; } for (int i = 0;i < n / 2;i++) { if (num[i] > 18 || num[i] < 0) return false; res[i] = (num[i] + 1) / 2 + '0'; res[n - i - 1] = num[i] / 2 + '0'; } return res[0] > '0'; } int main() { scanf("%s",s); int len = strlen(s); for (int i = 0;i < len;i++) num[i] = s[i] - '0'; if (check(len)) puts(res); else if (s[0] == '1' && len > 1) //为 “1”开头的数值进行特判 { for (int i = 0;i < len;i++) num[i] = s[i + 1] - '0'; len--; num[0] += 10; if (check(len)) puts(res); else puts("0"); } else puts("0"); return 0; }
Codeforces Round #342 (Div. 2) D. Finals in arithmetic(想法题/构造题)的更多相关文章
- Codeforces Round #342 (Div. 2) D. Finals in arithmetic 贪心
D. Finals in arithmetic 题目连接: http://www.codeforces.com/contest/625/problem/D Description Vitya is s ...
- Codeforces Round #342 (Div. 2) C. K-special Tables(想法题)
传送门 Description People do many crazy things to stand out in a crowd. Some of them dance, some learn ...
- Codeforces Round #342 (Div. 2)
贪心 A - Guest From the Past 先买塑料和先买玻璃两者取最大值 #include <bits/stdc++.h> typedef long long ll; int ...
- Codeforces Round #342 (Div. 2) C. K-special Tables 构造
C. K-special Tables 题目连接: http://www.codeforces.com/contest/625/problem/C Description People do many ...
- Codeforces Round #342 (Div. 2) B. War of the Corporations 贪心
B. War of the Corporations 题目连接: http://www.codeforces.com/contest/625/problem/B Description A long ...
- Codeforces Round #342 (Div. 2) A - Guest From the Past 数学
A. Guest From the Past 题目连接: http://www.codeforces.com/contest/625/problem/A Description Kolya Geras ...
- Codeforces Round #342 (Div. 2) E. Frog Fights set 模拟
E. Frog Fights 题目连接: http://www.codeforces.com/contest/625/problem/E Description stap Bender recentl ...
- Codeforces Round #342 (Div. 2) B. War of the Corporations(贪心)
传送门 Description A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Go ...
- Codeforces Round #342 (Div. 2) A. Guest From the Past(贪心)
传送门 Description Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the detai ...
随机推荐
- select、poll、epoll区别总结
1 本质上都是同步I/O 三者都是I/O复用,本质上都属于同步I/O.因为三者只是负责通知应用程序什么时候数据准备好了,实际的I/O操作还是在由应用程序处理:如果是异步I/O的话,实际I/O由内核处理 ...
- Java暗箱操作之自动装箱与拆箱
我以前在写Android项目的时候,估计写得最多最熟练的几句话就是: List<Integer> list = new ArrayList<Integer>(); list.a ...
- SqlServer--查询案例
use MyDataBase1 -- * 表示显示所有列 -- 查询语句没有加where条件表示查询所有行 select *from TblStudent ---只查询表中的部分列 select t ...
- jquery打造自定义控件(原创)
本人第一次发表文章,不足之出请大家多多包涵 下面是一个combox的代码 /// <reference path="../Js/jquery-1.7.2.min.js" /& ...
- 问题解决——MFC Ribbon 添加图标
=================================版权声明================================= 版权声明:本文为博主原创文章 未经许可不得转载 请通过右 ...
- 05.virsh命令的常用操作(kvm)
注:以下命令均可在virsh的man手册页中找到 KVM虚拟机管理常用命令(domain): virsh命令参数 功能 用法举例 list 查看已经存在的domain信息(可以带参数) vir ...
- Redis学习和环境搭建
基本的redis教程,搭建,可以参照下面任一教程: 地址一:http://www.yiibai.com/redis/redis_quick_guide.html 地址二:http://www.runo ...
- jquery——移动端滚动条插件iScroll.js
官网:http://cubiq.org/iscroll-5 demo: 滚动刷新:http://cubiq.org/dropbox/iscroll4/examples/pull-to-refresh/ ...
- 参数命令化 防止Sql注入
1.先在数据库中完成储存过程 create proc usp_selectStudent ) as select * froom student where StudentName=@studentN ...
- EhCache的配置
JPA和Hibernate的二级缓存都是这样做的 代码目录: 这是基础的jar包,如果少的话,再去maven下载 <!-- Spring --> <dependency> &l ...