A - Cinema Line

最新推荐文章于 2023-04-29 12:49:45 发布

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本文探讨了一种特定场景下的找零问题：电影院售票员如何确保每位顾客都能顺利购买到票价为25卢布的电影票，并获得正确的找零。文章提供了一段C++代码作为解决方案，该方案通过跟踪25卢布和50卢布钞票的数量来确定是否能够成功找零。

Description

The new "Die Hard" movie has just been released! There are n people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line?

Input

The first line contains integer n(1 ≤ n ≤ 10⁵) — the number of people in the line. The next line contains n integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line.

Output

Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO".

Sample Input

Input

4
25 25 50 50

Output

YES

Input

2
25 100

Output

NO

Input

4
50 50 25 25

Output

NO

题意：

排队给钱每人给25，看能不能找零钱，一开始收钱的人没有零钱

分析：

记录25元和50钱的个数，看能不能找零成功

代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    long int s,i,n,a[100005],q,w;
    while(cin>>n)
    {
        q=w=0;
        for(i=0;i<n;i++)
            cin>>a[i];
        for(i=0;i<n;i++)
        {
            if(a[i]==50&&q>0)
            {
                q--;
                w++;
            }
            else
                if(a[i]==100&&q&&w)
                {
                    q--;
                    w--;
                }
            else
                if(a[i]==100&&q>2)  q-=3;
             else
                if(a[i]==25)    q++;
            else
                break;

        }
        if(i==n)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
}

感受：

没有一次AC，忘了找100块钱还可以给三张25的