Codeforces Round #115 A. Robot Bicorn Attack 暴力
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
- A - Robot Bicorn Attack
Description Vasya plays Robot Bicorn Attack. The game consists of three rounds. For each one a non-n ...
- 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 ...
- 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 ...
- 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/ ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- <十一>面向对象分析之UML核心元素之组件
组件
- 配置ORACLE 客户端连接到数据库
--================================= -- 配置ORACLE 客户端连接到数据库 --================================= Oracle ...
- 【转】ios app 应用内购买配置完全指南
转自:http://blog.sina.com.cn/s/blog_4b55f6860100sbfb.html 第一印象觉得In-App Purchase(简称IAP)非常简单.Apple提供的大量文 ...
- hdu 1024(最大和连续子序列增强版)
题意:最大和连续子序列的增强版,要求从一序列中取出若干段,这些段之间不能交叉,使得和最大并输出. 分析:用dp[i][j]表示前j个数取出i段得到的最大值,那么状态转移方程为dp[i][j]=max( ...
- 六款最佳Linux教育应用
导读 对教育行业的用户来说,有好几款专门的Linux发行版是专门面向教育行业的.本文将介绍适合教育领域的几款顶级发行版. 1.Edubuntu 位居榜首的是Edubuntu.顾名思义,Edubuntu ...
- Delphi 使用串口模拟工具进行串口程序开发调试
版权声明:本文为博主原创文章,如需转载请注明出处及作者. 本文由小李专栏原创,转载需注明出处:[http://blog.csdn.net/softwave/article/details/8907 ...
- 【PHP】Windows环境Hello World
转自:http://www.cnblogs.com/wangkangluo1/archive/2011/07/19/2110943.html 一 下载 XAMPP下载地址: https://sourc ...
- 使用Qmake在树莓派上开发Opencv程序
Qt 安装 PC 端 下载安装即可 https://mirrors.ustc.edu.cn/qtproject/official_releases/qt 树莓派:Qt开发套件和opencv安装sud ...
- WeChat Official Account Admin Platform Message API Guide
Keyword: WeChat Message API Text Image Location Link Event Music RichMedia Author: PondBay Studio[We ...
- 链表逆序(JAVA实现)
题目:将一个有链表头的单向单链表逆序 分析: 链表为空或只有一个元素直接返回: 设置两个前后相邻的指针p,q,使得p指向的节点为q指向的节点的后继: 重复步骤2,直到q为空: 调整链表头和链表尾: 图 ...