UPC 6351 Lemonade Line

最新推荐文章于 2020-02-26 22:06:25 发布

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本文介绍了一道关于农场动物排队购买柠檬水的算法题。通过分析每头牛愿意等待的最牛数量来确定最少可能有多少头牛会加入队伍。文章提供了一个C++实现方案，采用排序算法来解决该问题。

Lemonade Line

时间限制: 1 Sec 内存限制: 128 MB

题目描述

It's a hot summer day out on the farm, and Farmer John is serving lemonade to his N cows! All N cows (conveniently numbered 1…N) like lemonade, but some of them like it more than others. In particular, cow ii is willing to wait in a line behind at most wiwi cows to get her lemonade. Right now all NN cows are in the fields, but as soon as Farmer John rings his cowbell, the cows will immediately descend upon FJ's lemonade stand. They will all arrive before he starts serving lemonade, but no two cows will arrive at the same time. Furthermore, when cow ii arrives, she will join the line if and only if there are at most wiwi cows already in line.
Farmer John wants to prepare some amount of lemonade in advance, but he does not want to be wasteful. The number of cows who join the line might depend on the order in which they arrive. Help him find the minimum possible number of cows who join the line.

输入

The first line contains N, and the second line contains the N space-separated integers w1,w2,…,wN. It is guaranteed that 1≤N≤105, and that 0≤wi≤109 for each cow i.

输出

Print the minimum possible number of cows who might join the line, among all possible orders in which the cows might arrive.

样例输入

5
7 1 400 2 2

样例输出

提示

In this setting, only three cows might end up in line (and this is the smallest possible). Suppose the cows with w=7 and w=400 arrive first and wait in line. Then the cow with w=1 arrives and turns away, since 2 cows are already in line. The cows with w=2 then arrive, one staying and one turning away.

题意：有N头牛，每头奶牛有一个值W，当这头牛到来时前面最多只能有W头牛在排队，它才会排到队伍后面，否则它就走了。农夫想要给最少的柠檬水，求最多有几头牛排队。

排序，把W值大的放前面，这样让后面牛不能排队

#include<algorithm>
#include<cstdio>
using namespace std;

typedef long long LL;

LL cmp(LL a, LL b){
    return a>b;
}
 
int main()
{
    int n;
    scanf("%d", &n);
    LL arr[n], ans = 0;
    for(int i = 0; i < n; i++)
        scanf("%lld",&arr[i]);
    sort(arr, arr+n, cmp);
    for(int i = 0; i < n; i++)        if(arr[i] >= ans) ans ++;
    printf("%lld\n", ans);
 
    return 0;
}