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
- HWM的实验
HWM是数据段中使用空间和未使用空间之间的界限,假如现有自由链表上的数据块不能满足需求,Oracle把HWM指向的数据块加入到自由链表上,HWM向前移动到下一个数据块.简单说,一个数据段中,HWM左边 ...
- C#中的lock关键字
前几天与同事激烈讨论了一下,有一点收获,记录起来. 首先给出MSDN的定义: lock 关键字可以用来确保代码块完成运行,而不会被其他线程中断.这是通过在代码块运行期间为给定对象获取互斥锁来实现的. ...
- 修改Oracle 表空间名称 tablespace name
修改表空间名称步骤如下: 1. 使用oracle用户登录执行 $sqlplus / as sysdba 2. 执行修改表空间命令如下 SQL> alter tablespace TEST re ...
- Corn Fields(POJ 3254状压dp)
题意: n*m网格1能放0不能放 放的格子不能相邻 求一共多少种可放的方案. 分析: dp[i][j]第i行可行状态j的的最大方案数,枚举当前行和前一行的所有状态转移就行了(不放牛也算一种情况) #i ...
- SQL经典笔试题之一
本题用到下面三个关系表: CARD 借书卡. CNO 卡号,NAME 姓名,CLASS 班级 BOOKS 图书. BNO 书号,BNAME 书名,AUTHOR 作者,PRIC ...
- strcasecmp在VS2010中提示未定义标识符
分析: strcasecmp(*,*)是用来比较字符串,定义在string.h头文件中,但是在windows下即使添加string.h头文件,依然会报错. 解决: 添加 #if defined(_MS ...
- Python对象体系揭秘
Guido用C语言创造了Python,在Python的世界中一切皆为对象. 一.C视角中的Python对象 让我们一起追溯到源头,Python由C语言实现,且向外提供了C的API http://doc ...
- 利用ASP.NET MVC源代码调试你的应用程序[转]
由于项目需要,最近学起asp.net mvc.昨天遇到ViewData和TempData他们之间的分别这样让我纠结的问题.有园友强烈建议我去看ASP.NET MVC的源代码.所以,我想到如何在调试AS ...
- Linux 中 x86 的内联汇编
工程中需要用到内联汇编,找到一篇不错的文章,趁机学习下. 原文地址:http://www.ibm.com/developerworks/cn/linux/sdk/assemble/inline/ 如果 ...
- 王家林的81门一站式云计算分布式大数据&移动互联网解决方案课程第14门课程:Android软硬整合设计与框架揭秘: HAL&Framework &Native Service &App&HTML5架构设计与实战开发
掌握Android从底层开发到框架整合技术到上层App开发及HTML5的全部技术: 一次彻底的Android架构.思想和实战技术的洗礼: 彻底掌握Andorid HAL.Android Runtime ...