Codeforces 1082 B. Vova and Trophies-有坑 (Educational Codeforces Round 55 (Rated for Div. 2))
2 seconds
256 megabytes
standard input
standard output
Vova has won nn trophies in different competitions. Each trophy is either golden or silver. The trophies are arranged in a row.
The beauty of the arrangement is the length of the longest subsegment consisting of golden trophies. Vova wants to swap two trophies (not necessarily adjacent ones) to make the arrangement as beautiful as possible — that means, to maximize the length of the longest such subsegment.
Help Vova! Tell him the maximum possible beauty of the arrangement if he is allowed to do at most one swap.
The first line contains one integer nn (2≤n≤1052≤n≤105) — the number of trophies.
The second line contains nn characters, each of them is either G or S. If the ii-th character is G, then the ii-th trophy is a golden one, otherwise it's a silver trophy.
Print the maximum possible length of a subsegment of golden trophies, if Vova is allowed to do at most one swap.
10
GGGSGGGSGG
7
4
GGGG
4
3
SSS
0
In the first example Vova has to swap trophies with indices 44 and 1010. Thus he will obtain the sequence "GGGGGGGSGS", the length of the longest subsegment of golden trophies is 77.
In the second example Vova can make no swaps at all. The length of the longest subsegment of golden trophies in the sequence is 44.
In the third example Vova cannot do anything to make the length of the longest subsegment of golden trophies in the sequence greater than 00.
在左右两侧都是G,中间为S的时候,在最后需要和直接连续G的长度交换一个S为G的比较一下,wa在这里了。
代码:
//B
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+;
const int inf=0x3f3f3f3f; char c[maxn];
vector<int> ve; int main()
{
int n;cin>>n;
cin>>c;
int num=;
for(int i=;i<n;i++){
if(c[i]=='G') num++;
if(c[i-]=='G'&&c[i]=='S'&&c[i+]=='G') ve.push_back(i);
}
int ret=,tmp=;
for(int i=;i<n;i++){
if(c[i]=='G') tmp++;
else ret=max(ret,tmp),tmp=;
}
ret=max(ret,tmp);
ret=min(ret+,num);
if(num==n) {cout<<n<<endl;return ;}
if(num==) {cout<<<<endl;return ;}
if(ve.size()==){
int ans=-,cnt=;
for(int i=;i<n;i++){
if(c[i]=='G') cnt++;
if(c[i]=='S') ans=max(ans,cnt),cnt=;
}
ans=max(ans,cnt);
cout<<min(num,ans+)<<endl;
}
else{
int ans=,flag=;
for(auto it:ve){
int cnt=;
for(int i=it-;i>=;i--){
if(c[i]=='S') break;
else cnt++;
}
for(int i=it+;i<n;i++){
if(c[i]=='S') break;
else cnt++;
}
ans=max(ans,cnt);
}
ans=min(num,ans+);
cout<<max(ans,ret)<<endl;
}
} /*
16
GSGSSGSSGGGSSSGS 4
*/
Codeforces 1082 B. Vova and Trophies-有坑 (Educational Codeforces Round 55 (Rated for Div. 2))的更多相关文章
- Educational Codeforces Round 55 (Rated for Div. 2) B. Vova and Trophies 【贪心 】
传送门:http://codeforces.com/contest/1082/problem/B B. Vova and Trophies time limit per test 2 seconds ...
- Educational Codeforces Round 55 (Rated for Div. 2) B. Vova and Trophies (贪心+字符串)
B. Vova and Trophies time limit per test2 seconds memory limit per test256 megabytes inputstandard i ...
- Educational Codeforces Round 55 (Rated for Div. 2) Solution
A. Vasya and Book Solved. 三种方式取$Min$ #include <bits/stdc++.h> using namespace std; #define ll ...
- Educational Codeforces Round 55 (Rated for Div. 2) B. Vova and Trophies
传送门 https://www.cnblogs.com/violet-acmer/p/10035971.html 题意: Vova有n个奖杯,这n个奖杯全部是金奖或银奖,Vova将所有奖杯排成一排,你 ...
- Codeforces 1082 C. Multi-Subject Competition-有点意思 (Educational Codeforces Round 55 (Rated for Div. 2))
C. Multi-Subject Competition time limit per test 2 seconds memory limit per test 256 megabytes input ...
- Codeforces 1082 D. Maximum Diameter Graph-树的直径-最长链-构造题 (Educational Codeforces Round 55 (Rated for Div. 2))
D. Maximum Diameter Graph time limit per test 2 seconds memory limit per test 256 megabytes input st ...
- Codeforces 1082 A. Vasya and Book-题意 (Educational Codeforces Round 55 (Rated for Div. 2))
A. Vasya and Book time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...
- Educational Codeforces Round 55 (Rated for Div. 2) A/B/C/D
http://codeforces.com/contest/1082/problem/A WA数发,因为默认为x<y = = 分情况讨论,直达 or x->1->y or x-& ...
- Educational Codeforces Round 55 (Rated for Div. 2) C. Multi-Subject Competition 【vector 预处理优化】
传送门:http://codeforces.com/contest/1082/problem/C C. Multi-Subject Competition time limit per test 2 ...
随机推荐
- vijos 1066 弱弱的战壕 树状数组
描述 永恒和mx正在玩一个即时战略游戏,名字嘛~~~~~~恕本人记性不好,忘了-_-b. mx在他的基地附近建立了n个战壕,每个战壕都是一个独立的作战单位,射程可以达到无限(“mx不赢定了?!?”永恒 ...
- R、Python、Scala和Java,到底该使用哪一种大数据编程语言?
有一个大数据项目,你知道问题领域(problem domain),也知道使用什么基础设施,甚至可能已决定使用哪种框架来处理所有这些数据,但是有一个决定迟迟未能做出:我该选择哪种语言?(或者可能更有针对 ...
- ES6新用法
ES6 详细参考页面 简介 ECMAScript和JavaScript的关系是,前者是后者的规格,后者是前者的一种实现.一般来说,这两个词是可以互换的. let命令 ES6新增了let命令,用来声明变 ...
- sass_sass安装
你会不会因为有些事遇到各种各样的问题而搁置,直到把这个事情被耽误了几天.最近一直在弄sass这个东西,安装的过程中各种问题.sass是一个基于ruby环境开发的,安装sass之前得先把ruby给安装了 ...
- D题 hdu 1412 {A} + {B}
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1412 {A} + {B} Time Limit: 10000/5000 MS (Java/Others ...
- linux非阻塞的socket EAGAIN的错误处理【转】
转自:http://blog.csdn.net/tianmohust/article/details/8691644 版权声明:本文为博主原创文章,未经博主允许不得转载. 在Linux中使用非阻塞的s ...
- Win7蓝屏代码0X0000007B可能是SATA mode问题
Win7蓝屏代码0X0000007B可能是硬盘模式的问题,我进入BIOS把SATA的mode从Enhanced改为Compatible(及IDE兼容模式)结果系统可以顺利启动没有问题. 从 ...
- [New learn] 设计模式
本文翻译自:http://www.raywenderlich.com/46988/ios-design-patterns iOS设计模式 - 你可能听到过这个术语,但是你知道是什么意思吗?虽然大多数的 ...
- [ Python ] 基本数据类型及属性(上篇)
1. 基本数据类型 (1) 数字 - int (2) 字符串 - str (3) 布尔值 - bool 2. int 类型中重要的方法 (1) int 将字符串转 ...
- lnmp的安装--php
1.下载php源码 wget http://cn2.php.net/distributions/php-5.6.3.tar.gz tar zxvf php-5.6.3.tar.gz cd php-5. ...