Codeforces Round #374 (Div. 2) A. One-dimensional Japanese Crossword —— 基础题

本文介绍了一道关于一维日本填字游戏的问题,并提供了一个简洁高效的解决方案。该问题要求根据一行黑白方格的数据,输出连续黑色方格组的数量及其长度。

题目链接:http://codeforces.com/contest/721/problem/A


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.



题解:

虽然是一道很基础的题目,但是却清晰了我对类似情况的写法, 值得记录一下。

关键代码:

if(a[i]=='B')
{
      if(i==0 || a[i-1]=='W')  k++;         
      s[k]++;
}


代码如下:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <sstream>
#include <algorithm>
using namespace std;
#define ms(a, b)  memset((a), (b), sizeof(a))
#define eps 0.0000001
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int maxn = 100+10;

int n;
char a[maxn];
int k, s[maxn];

void init()
{
    scanf("%d",&n);
    scanf("%s",a);
    ms(s,0);
    k = 0;
}

void solve()
{
    int cnt = 1;
    for(int i = 0; i<n; i++)
    {
        if(a[i]=='B')
        {
            if(i==0 || a[i-1]=='W')  k++;
            s[k]++;
        }
    }

    printf("%d\n",k);
    for(int i = 1; i<=k; i++)
        printf("%d ",s[i]);
    putchar('\n');
}

int main()
{
//    int T;
//    scanf("%d",&T);
//    while(T--)
    {
        init();
        solve();
    }
    return 0;
}


转载于:https://www.cnblogs.com/DOLFAMINGO/p/7538698.html

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