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  • CF Taxi(贪心)

    B. Taxi
    time limit per test
    3 seconds
    memory limit per test
    256 megabytes

    input
    standard input
    output
    standard output

    After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?

    Input

    The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.

    Output

    Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.

    Examples
    input
    Copy
    5
    1 2 4 3 3
    
    output
    Copy
    4
    
    input
    Copy
    8
    2 3 4 4 2 1 3 1
    
    output
    Copy
    5
    
    Note

    In the first test we can sort the children into four cars like this:

    • the third group (consisting of four children),
    • the fourth group (consisting of three children),
    • the fifth group (consisting of three children),
    • the first and the second group (consisting of one and two children, correspondingly).

    There are other ways to sort the groups into four cars.

    思路:四个人的小组做一辆车,三个人的小组坐一辆车,每两个两个人的小组做一辆车,一个人的小组分别与两个人的小组和三个人的小组合作一辆车,剩余的一个人的小组每四个人一辆车。

    过题代码:

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <functional>
    #include <iostream>
    using namespace std;
    int a[5];
    int main(){
        int n,x,ans=0;
        scanf("%d",&n);
        for (int i=1;i<=n;i++){
            scanf("%d",&x);
            a[x]++;
        }
        ans+=a[4]+a[3]+a[2]/2;
        if(a[2]%2) ans++,a[1]-=2>=a[1]?a[1]:2;
        a[1]-=a[3]>a[1]?a[1]:a[3];
        if(a[1]%4) ans++;
        ans+=a[1]/4;
        printf("%d
    ",ans);
        return 0;
    
    }
    

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  • 原文地址:https://www.cnblogs.com/acerkoo/p/9490342.html
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