牛客多校第三场A【Clam and fish】贪心

最新推荐文章于 2020-08-07 15:08:02 发布

原创最新推荐文章于 2020-08-07 15:08:02 发布 · 328 阅读

0 ·

CC 4.0 BY-SA版权

牛客多校专栏收录该内容

6 篇文章

订阅专栏

本文深入解析了牛客网上的Clamandfish贪心算法问题，通过详细解释游戏规则和策略，展示了如何计算在给定阶段类型下能够捕获的最大鱼数。文章提供了代码实现，并附带了示例输入输出，帮助读者理解算法的运作过程。

A【Clam and fish】贪心

链接：https://ac.nowcoder.com/acm/contest/5668/A
来源：牛客网
题目：
There is a fishing game as following:

The game contains nn stages, numbered from 11 to nn.

There are four types of stages (numbered from 00 to 33):

type 00: There are no fish and no clam in this stage.

type 11: There are no fish and one clam in this stage.

type 22: There are one fish and no clam in this stage.

type 33: There are one fish and one clam in this stage.

In each stage, you can do exactly one of the following four actions.

If there is a clam in the stage, you can use this clam to make one pack of fish bait. And the number of packs of fish bait you have is increased by one. You can use this pack of fish bait to catch fish after this stage.

If there is one fish in the stage, you can catch this fish without any fish bait. After this stage, the number of packs of fish bait you have is not changed.

If you have at least one pack of fish bait. You can always catch one fish by using exactly one pack of fish bait even if there are no fish in this stage. After this stage, the number of packs of fish bait you have is decreased by one.

You can do nothing.

Now, you are given nn and the type of each stage. Please calculate the largest number of fish you can get in the fishing game.

输入描述:
The first line contains one integer t (1 \le t \le 2.5 \times 10^51≤t≤2.5×10
5
) — the number of test cases.

There are two lines in each test. The first line contains one integer n (1 \le n \le 2 \times 10^61≤n≤2×10
6
), indicating the number of stages in this game. The second line contains a string with length n. The i-th character of this string indicates the type of the i-th stage.

The sum of n across the test cases doesn’t exceed 2 \times 10^62×10
6
.
输出描述:
For each test case print exactly one integer — the maximum number of fish you can catch in the game configuration.
示例1

输入
2
4
0103
1
1
输出
2
0
备注:
One possible scenario you can catch two fishes is as follow:

stage 1: Do nothing.
stage 2: Make a pack of fish bait.
stage 3: Catch a fish by a pack of fish bait.
stage 4: Catch the fish that appears in the stage.

总结：有鱼就抓（之前看直播讲题的时候这句话印象超深刻）
就是2 3这两种情况，直接fish++就好，碰到0，有饵就fish++，碰到1就bait++。最后sum=fish+bait/2;（/2的意义就是：举个例子，11两次，一次制饵一次钓鱼是获得🐟最多的方式，所以以此类推便可得/2）
最开始写的是自定义函数不知道为什么过不了，要是有大佬用自定义函数过了麻烦告诉我一波。

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<stack>
#include<algorithm>
using namespace std;

typedef long long ll;


int main()
{
    ll k;
    cin>>k;
    while(k--)
    {
        ll x;
        char s[200006];
        scanf("%lld%s",&x,s);
        ll sum=0,fish=0,bait=0;
        for(int i=0;i<x;i++)
        {

            if(s[i]=='0')
                {
                if(bait>0)
                    {
                        fish++;
                        bait--;
                    }
                }
            else if(s[i]=='1')
            {
                bait++;
            }
            else fish++;
            //cout<<fish;
        }
        sum+=fish+bait/2;
        cout<<sum<<endl;
    }
   return 0;

}