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  • A. Cinema Line

    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    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".

    Examples
    input
    Copy
    4
    25 25 50 50
    output
    Copy
    YES
    input
    Copy
    2
    25 100
    output
    Copy
    NO
    input
    Copy
    4
    50 50 25 25
    output
    Copy
    NO


    #include<iostream>
    #include<string.h>
    using namespace std;
    int main()
    {
        int n,a[100005];
        cin>>n;
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
        }
        int t1=0,t2=0,flag=1;
        for(int i=0;i<n;i++)
        {
            if(a[i]==25)
            {
                t1++;
                continue;
            }
            if(a[i]==50)
            {
                if(t1>=1)
                {
                    t2++;
                    t1--;
                    continue;
                }
                else
                {
                    flag=0;
                    break;
                }
            }
            if(a[i]==100)
            {
                if(t1>=1&&t2>=1)
                {
                    t1--;
                    t2--;
                    continue;
                }
                if(t1>=3)
                {
                    t1=t1-3;
                    continue;
                }
    
                else
                {
                    flag=0;
                    break;
                }
            }
    
        }
        if(flag==1)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/-citywall123/p/9703623.html
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