Educational Codeforces Round 42 (Rated for Div. 2) B

本文转载自 buerdepepeqi 查看原文 2018/04/11 15 div/ code/ for/ codeforces

There are $n$

The university team for the Olympiad consists of $a$

no student-programmer is sitting next to the student-programmer;
and no student-athlete is sitting next to the student-athlete.

In the other words, there should not be two consecutive (adjacent) places where two student-athletes or two student-programmers are sitting.

Consider that initially occupied seat places are occupied by jury members (who obviously are not students at all).

Input

The first line contain three integers $n$

The second line contains a string with length $n$

Output

Print the largest number of students, which you can put in the railway carriage so that no student-programmer is sitting next to a student-programmer and no student-athlete is sitting next to a student-athlete.

Examples

input

Copy

5 1 1
*...*

output

Copy

input

Copy

6 2 3
*...*.

output

Copy

input

Copy

11 3 10
.*....**.*.

output

Copy

input

Copy

3 2 3
***

output

Copy

Note

In the first example you can put all student, for example, in the following way: *.AB*

In the second example you can put four students, for example, in the following way: *BAB*B

In the third example you can put seven students, for example, in the following way: B*ABAB**A*B

The letter A means a student-programmer, and the letter B — student-athlete.

题意：在一排座位上，有空位，有两种类型的人需要我们去给他们排座位，相同类型的人不能坐在一起，问最多可以坐多少个人

我觉得这是一个贪心模拟题，min(可用长度,a的人数) + min(b,长度) +min(（剩余a+b) ,额外座位数)，一开始wa了好多发是因为没有判断剩余的a和b哪个更加大、

所以这题的几个坑点是：1.从左到右每次排座位时要注意当前位置的左边

　　　　　　　　　　　2.要想坐最多的人就应该先排人数多的那一种类型

　　　　　　　　　　　3.要判断当前的a类型或b类型是否为0，提前判断

附上代码

#include <bits/stdc++.h>
using namespace std;
char s[200005];
int main(){
    int n, a, b;
    scanf("%d%d%d", &n, &a, &b);
    scanf("%s", s+1);
    int l = strlen(s+1);
    s[0] = '.';
    int ans = 0;
    for(int i=1;i<=l;i++){
        int mx = max(a, b);
        if(s[i] == '*') continue;
        if(a == 0 && b == 0) break;
        if(a == 0){
            if(s[i-1] == 'b') continue;
            s[i] = 'b';
            b--;
            ans++;
        }
        else if(b == 0){
            if(s[i-1] == 'a') continue;
            s[i] = 'a';
            a--;
            ans++;
        }
        else{
            if(mx == a){
                if(s[i-1] == 'a'){
                    s[i] = 'b';
                    b--;
                }
                else{
                    s[i] = 'a';
                    a--;
                }
                ans++;
            }
            else if(mx == b){
                if(s[i-1] == 'b'){
                    s[i] = 'a';
                    a--;
                }
                else{
                    s[i] = 'b';
                    b--;
                }
                ans++;
            }
        }
    }
    printf("%d\n", ans);
}

View Code

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