CF#202DIV2: A. Cinema Line

 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 single100,50 or 25 ruble bill. A "Die Hard" ticket costs25 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 containsnintegers, each of them equals25, 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 test(s)

Input

4

25 25 50 50

Output

YES

Input

2

25 100

Output

NO

Input

4

50 50 25 25

Output

NO

 

题意：只有25,50,100三种钱币，一开始售票员没其他钱币，靠收到的钱来找零，票价25，给出买票人给钱的顺序，问能否将所有人找零

 

CF第一题照例是超级水题

 

 

 

 

#include <stdio.h>  
#include <string.h>  
#include <algorithm>  
using namespace std;  
  
int hash[105];  
  
int main()  
{  
    int n,i,a;  
    while(~scanf("%d",&n))  
    {  
        int flag = 0;  
        memset(hash,0,sizeof(hash));  
        for(i = 0; i<n; i++)  
        {  
            scanf("%d",&a);  
            hash[a]++;  
            if(a == 25)  
                continue;  
            else if(a == 50)  
            {  
                if(hash[25])  
                {  
                    hash[25]--;  
                    continue;  
                }  
                else  
                    flag = 1;  
            }  
            else if(a == 100)  
            {  
                if(hash[25] && hash[50])  
                {  
                    hash[25]--;  
                    hash[50]--;  
                    continue;  
                }  
                else if(hash[25]>=3)  
                {  
                    hash[25]-=3;  
                    continue;  
                }  
                else  
                    flag = 1;  
            }  
        }  
        if(flag)  
            printf("NO\n");  
        else  
            printf("YES\n");  
    }  
  
    return 0;  
}  

 

 

 

补充：软件开发 , C++ ,