2016 Winter Training Day #1_G题_codefcrces 349A(贪心)

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 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

425 25 50 50

output

YES

input

225 100

output

NO

input

450 50 25 25

output

NO

题意：有一个销售员，和n个买电影票的人，货币有三种金额：25,50,100.  每个人都拿着这三种金额中的一种去买票，问售货员能否给每个人都成功找零。

思路：简单贪心，每次找零都优先给客人50元的（金额尽可能大），这样才能使剩下的金额数量一样，但是灵活性更强，因为如果剩下全是50，100的话，就无法再找零了。

第一次wa是因为如果50没有了的时候，25的数量应该是 x -= 3的（75），但我写成了 x -= 2（50）;

 代码：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>using namespace std;const int INF = 0x7fffffff;int a[100500];int main(){    int n;    cin >> n;    int x = 0, y = 0, z = 0;    for(int i = 0; i < n; ++i)    {        cin >> a[i];    }    for(int i = 0; i < n; ++i)    {        if(a[i] == 25)            x++;        else if(a[i] == 50)        {            x--;            y++;        }        else        {            if(y)            {                z++;                x--;                y--;            }            else            {                z++;                x-=3;            }        }        if(x < 0 || y < 0 || z < 0)        {            cout << "NO" << endl;            return 0;        }    }    cout << "YES" << endl;    return 0;}