A. One-dimensional Japanese Crossword

time limit per test:1 second
memory limit per test:256 megabytes
input:standard input
output:standard output

Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized a × b squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia https://en.wikipedia.org/wiki/Japanese_crossword).

Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of n squares (e.g. japanese crossword sized 1 × n), which he wants to encrypt in the same way as in japanese crossword.

The example of encrypting of a single row of japanese crossword.

Help Adaltik find the numbers encrypting the row he drew.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100) — the length of the row. The second line of the input contains a single string consisting of n characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).

Output

The first line should contain a single integer k — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.

The second line should contain k integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.

Examples

input

3
BBW

output

1
2

input

5
BWBWB

output

3
1 1 1

input

4
WWWW

output

0

input

4
BBBB

output

1
4

input

13
WBBBBWWBWBBBW

output

3
4 1 3

Note

The last sample case correspond to the picture in the statement.

第一次打cf,签到题,问黑色块的个数和长度

 //2016.9.30
#include <iostream>
#include <cstdio>
#include <cstring>
#define N 105 using namespace std; string str;
int ans[N]; int main()
{
int n, cnt, tmp;
while(scanf("%d", &n)!=EOF)
{
tmp = cnt = ;
cin>>str;
int len = str.length();
for(int i = ; i < len; i++)
{
if(str[i]=='B')
{
tmp = ;
while(str[i]=='B'&&i<n)
{
i++;
tmp++;
}
ans[cnt++] = tmp;
}
}
cout<<cnt<<endl;
for(int i = ; i < cnt; i++)
{
if(i==)cout<<ans[i];
else cout<<" "<<ans[i];
}
if(!cnt)cout<<endl;
}
return ;
}

CodeForces 721A的更多相关文章

  1. CodeForces 721A One-dimensional Japanese Crossword (水题)

    题意:给定一行字符串,让你输出字符‘B'连续出现的次数. 析:直接扫一下就OK了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024 ...

  2. 【76.57%】【codeforces 721A】One-dimensional Japanese Crossword

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  3. Codeforces水题集合[14/未完待续]

    Codeforces Round #371 (Div. 2) A. Meeting of Old Friends |B. Filya and Homework A. Meeting of Old Fr ...

  4. python爬虫学习(5) —— 扒一下codeforces题面

    上一次我们拿学校的URP做了个小小的demo.... 其实我们还可以把每个学生的证件照爬下来做成一个证件照校花校草评比 另外也可以写一个物理实验自动选课... 但是出于多种原因,,还是绕开这些敏感话题 ...

  5. 【Codeforces 738D】Sea Battle(贪心)

    http://codeforces.com/contest/738/problem/D Galya is playing one-dimensional Sea Battle on a 1 × n g ...

  6. 【Codeforces 738C】Road to Cinema

    http://codeforces.com/contest/738/problem/C Vasya is currently at a car rental service, and he wants ...

  7. 【Codeforces 738A】Interview with Oleg

    http://codeforces.com/contest/738/problem/A Polycarp has interviewed Oleg and has written the interv ...

  8. CodeForces - 662A Gambling Nim

    http://codeforces.com/problemset/problem/662/A 题目大意: 给定n(n <= 500000)张卡片,每张卡片的两个面都写有数字,每个面都有0.5的概 ...

  9. CodeForces - 274B Zero Tree

    http://codeforces.com/problemset/problem/274/B 题目大意: 给定你一颗树,每个点上有权值. 现在你每次取出这颗树的一颗子树(即点集和边集均是原图的子集的连 ...

随机推荐

  1. openssl windows编译 32位&64位

    openssl版本:openssl-1.0.0k 64位编译 1.编译环境: openssl-1.0.0a必须用vs2008编译(Open Visual Studio 2008 x64 Cross T ...

  2. css3盒模型学习--利用box自适应布局

    box-flex是css3新添加的盒子模型属性,它的出现可以解决我们通过N多结构.css实现的布局方式.经典   的一个布局应用就是布局的垂直等高.水平均分.按比例划分. 目前box-flex属性还没 ...

  3. [Lua]Mac系统上安装Lua环境

    1.下载 Lua语言的官方网站 http://www.lua.org/ 下载最新版本的Lua环境 2.安装 解压下载包lua-5.3.1.tar.gz 打开终端Terminal 使用cd命令进入该目录 ...

  4. K-means算法简介

    K-means 算法是无监督的 聚类算法,算法简单,有效. K-means算法: 输入参数: 指定聚类数目 k,训练集 X 输出 : k 个聚类 算法描述: K-means 算法 是一个 迭代算法,每 ...

  5. POJ 2395 Out of Hay

    这个问题等价于求最小生成树中权值最大的边. #include<cstdio> #include<cstring> #include<cmath> #include& ...

  6. 19、手把手教你Extjs5(十九)模块Grid的其他功能的设想

    经过对自定义模块和Grid的设计和编码,现在已经能对一个有配置信息的模块来生成界面并进行一些简单的CURD操作.由于这是一个全解释性的前台的架构,因此你想到的任何新主意都可以放到所有的模块中. 比如对 ...

  7. ubuntu apache2 ssl配置

    Ubuntu下HTTPS配置非常简单,对大部分用户而言,使用普通的自签名证书,只需按照步骤进行就可以了,无需了解密钥.证书的更多知识,更深的背景 知识还有RSA算法.DES算法.X509规范.CA机构 ...

  8. FreeRTOS基础以及UIP之协程--C语言剑走偏锋

    在FreeRTOS中和UIP中,都使用到了一种C语言实现的多任务计数,专业的定义叫做协程(coroutine),顾名思义,这是一种协作的例程, 跟具有操作系统概念的线程不一样,协程是在用户空间利用程序 ...

  9. Python字符串的encode与decode研究心得——解决乱码问题

    转~Python字符串的encode与decode研究心得——解决乱码问题 为什么Python使用过程中会出现各式各样的乱码问题,明明是中文字符却显示成“/xe4/xb8/xad/xe6/x96/x8 ...

  10. 《C程序设计语言》读书笔记----习题1-20

    练习1-20:编写程序detab,将输入中的制表符替换成适当数目的空格,使得空格充满到下一个制表符终止位的地方,.假设制表符终止位的位置时固定的,比如每隔n列就会出现一个终止位. 这里要理解“制表符” ...