A. Robot Bicorn Attack

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/175/problem/A

Description

Vasya plays Robot Bicorn Attack.

The game consists of three rounds. For each one a non-negative integer amount of points is given. The result of the game is the sum of obtained points. Vasya has already played three rounds and wrote obtained points one by one (without leading zeros) into the string s. Vasya decided to brag about his achievement to the friends. However, he has forgotten how many points he got for each round. The only thing he remembers is the string s.

Help Vasya to find out what is the maximum amount of points he could get. Take into account that Vasya played Robot Bicorn Attack for the first time, so he could not get more than 1000000 (106) points for one round.

Input

The only line of input contains non-empty string s obtained by Vasya. The string consists of digits only. The string length does not exceed 30 characters.

Output

Print the only number — the maximum amount of points Vasya could get. If Vasya is wrong and the string could not be obtained according to the rules then output number -1.

Sample Input

1234

Sample Output

37

HINT

题意

给你一个字符串,要求使得字符串分为3节,并且每节都不能带首位0

并且每一节的数字都不能超过1e6

问你这三节的和最大可能为多少

题解:

暴力咯,数据范围只有30~

不过这道题如果数据范围出到1e5的话,是一道非常好玩的题目哦~

代码:

#include<iostream>
#include<stdio.h>
using namespace std; string s;
long long ans = ;
long long solve(int l,int r)
{
if(l!=r&&s[l]=='')return -;
long long res = ;
for(int i=l;i<=r;i++)
{
res = res * + (s[i]-'');
if(res>)
return -;
}
return res;
}
int main()
{
cin>>s;
int flag = ;
for(int i=;i<s.size();i++)
{
for(int j=i+;j<s.size();j++)
{
long long sum1 = solve(,i-);
long long sum2 = solve(i,j-);
long long sum3 = solve(j,s.size()-);
if(sum1==-||sum2==-||sum3==-)
continue;
ans = max(ans,sum1+sum2+sum3);
flag = ;
}
}
if(flag==)
return puts("-1");
printf("%d\n",ans);
}

Codeforces Round #115 A. Robot Bicorn Attack 暴力的更多相关文章

  1. A - Robot Bicorn Attack

    Description Vasya plays Robot Bicorn Attack. The game consists of three rounds. For each one a non-n ...

  2. Codeforces Round #307 (Div. 2) B. ZgukistringZ 暴力

    B. ZgukistringZ Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/551/probl ...

  3. Codeforces Round #115 B. Plane of Tanks: Pro 水题

    B. Plane of Tanks: Pro Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/17 ...

  4. Codeforces Round #328 (Div. 2) A. PawnChess 暴力

    A. PawnChess Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/592/problem/ ...

  5. Codeforces Round #404 (Div. 2)(A.水,暴力,B,排序,贪心)

    A. Anton and Polyhedrons time limit per test:2 seconds memory limit per test:256 megabytes input:sta ...

  6. Codeforces Round #369 (Div. 2) A B 暴力 模拟

    A. Bus to Udayland time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  7. Codeforces Educational Codeforces Round 5 B. Dinner with Emma 暴力

    B. Dinner with Emma 题目连接: http://www.codeforces.com/contest/616/problem/A Description Jack decides t ...

  8. Codeforces Round #188 (Div. 1) B. Ants 暴力

    B. Ants Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/317/problem/B Des ...

  9. Codeforces Round #329 (Div. 2) A. 2Char 暴力

    A. 2Char Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/593/problem/A De ...

随机推荐

  1. 学习Mongodb(一)

    图片摘录自陈彦铭出品2012.5的<10天掌握MongDB> MongoDB的特点--->面向集合存储,易于存储对象类型的数据--->模式自由--->支持动态查询---& ...

  2. 【Mysql】命令行

    查看数据库->show databases; 建数据库->create database 数据库名; 建表->use 数据库名;->create table 表名(字段); 查 ...

  3. 【编程之美】计算1-N中含1的个数

    转自:点我  1位数的情况: 在解法二中已经分析过,大于等于1的时候,有1个,小于1就没有. 2位数的情况: N=13,个位数出现的1的次数为2,分别为1和11,十位数出现1的次数为4,分别为10,1 ...

  4. PagerSlidingTabStrip 高亮选中标题

    1.选中标题后,高亮标题@Override public void onPageSelected(int position) { setSelectTextColor(position); if (d ...

  5. Redis Sentinel机制与用法

    概述 Redis-Sentinel是Redis官方推荐的高可用性(HA)解决方案,当用Redis做Master-slave的高可用方案时,假如master宕机了,Redis本身(包括它的很多客户端)都 ...

  6. C# 邮件发送系统

    using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; usin ...

  7. 《Python 学习手册4th》 第十七章 作用域

    ''' 时间: 9月5日 - 9月30日 要求: 1. 书本内容总结归纳,整理在博客园笔记上传 2. 完成所有课后习题 注:“#” 后加的是备注内容 (每天看42页内容,可以保证月底看完此书) “重点 ...

  8. pcduino+opencv实现人脸追踪摄像头

    Pcduino是一款兼容Arduino接口的mini pc,A8架构1Ghz的CPU,计算能力不俗,用来跑OpenCV刚刚好.这里就用他们实现一个可以跟随人脸移动的摄像头. 硬件清单: 1.Pcdui ...

  9. 45度地图遮挡问题解决方案(cocos2d-x)

    最近一直在做45度斜视角游戏,也就是isometric等容地图,俗称2.5D.地图上物体的前后遮挡是我遇到的第一个问题,总结一下处理方法. 遮挡问题(不知道术语),就是比如一个角色站在树后面,那么树要 ...

  10. bzoj 4448 [Scoi2015]情报传递(主席树,LCA)

    [题目链接] http://www.lydsy.com/JudgeOnline/problem.php?id=4448 [题意] 给定一颗树,询问一条路径上权值小于t-c的点数. [思路] 将一个2查 ...