【CodeForces - 349A】Cinema Line （貪心(其實不是貪心)，亂搞）
題幹：
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
4 25 25 50 50
Output
YES
Input
2 25 100
Output
NO
Input
4 50 50 25 25
Output
NO
題目大意：
就是說n個人排隊來買票，票價25元，每個人都出 25,50,100中的一種鈔票，售票處需要找零錢，，問能不能找錢成功。。
解題報告：
分情況亂搞就可以了。
AC程式碼：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
int bk[500];
int main()
{
int n,tmp,flag = 1;
cin>>n;
for(int i = 1; i<=n; i++) {
scanf("%d",&tmp);
bk[tmp]++;
if(flag == 0) continue;
if(tmp == 25) continue;
if(tmp == 50) {
if(bk[25] != 0) {
bk[25]--;continue;
}
}
if(tmp == 100) {
if(bk[50]>=1 && bk[25]>=1) {
bk[50]--;bk[25]--;
continue;
}
if(bk[25]>=3) {
bk[25]-=3;
continue;
}
}
flag = 0;
}
if(flag) puts("YES");
else puts("NO");
return 0 ;
}