HDU3348-coins

“Yakexi, this is the best age!” Dong MW works hard and get high pay, he has many 1 Jiao and 5 Jiao banknotes(纸币), some day he went to a bank and changes part of his money into 1 Yuan, 5 Yuan, 10 Yuan.(1 Yuan = 10 Jiao)
“Thanks to the best age, I can buy many things!” Now Dong MW has a book to buy, it costs P Jiao. He wonders how many banknotes at least,and how many banknotes at most he can use to buy this nice book. Dong MW is a bit strange, he doesn’t like to get the change, that is, he will give the bookseller exactly P Jiao.
Input
T(T<=100) in the first line, indicating the case number.
T lines with 6 integers each:
P a1 a5 a10 a50 a100
ai means number of i-Jiao banknotes.
All integers are smaller than 1000000.
Output
Two integers A,B for each case, A is the fewest number of banknotes to buy the book exactly, and B is the largest number to buy exactly.If Dong MW can’t buy the book with no change, output “-1 -1”.
Sample Input
3
33 6 6 6 6 6
10 10 10 10 10 10
11 0 1 20 20 20
Sample Output
6 9
1 10
-1 -1

分析：

题意：
硬币有1元、5元、10元、50元、100元各若干，要买一件价格为P的商品，求所花硬币数量最少和最多各是多少？

解析：
最少是多少好弄，贪心算法嘛！但是最多怎么弄？这里我们就要换个思路了！所花的硬币最多，则剩下的银币最少，这还是贪心！这就是两次贪心嘛！

代码：

#include<iostream>
#include<algorithm> 
#include<cstdio>

using namespace std;

int book[10];
int money[5]={1,5,10,50,100};

int main()
{
	int T,p,sum,Minum,Manum,num;
	scanf("%d",&T);
	while(T--)
	{
		Minum=0,Manum=0,num=0;
		scanf("%d",&p);
		sum=p;
		for(int i=1;i<=5;i++)
		{
			scanf("%d",&book[i]);
		}
		int n;
		for(int i=5;i>0;i--)
		{
			if(sum<money[i-1]||!book[i])
				continue;
			n=min(sum/money[i-1],book[i]);
			sum-=money[i-1]*n;
			Minum+=n;
		}
		if(sum!=0)
			printf("-1 -1\n");
		else
		{
			for(int i=1;i<=5;i++)
			{
				sum+=book[i]*money[i-1];
				num+=book[i];
			}
			sum-=p;
			for(int i=5;i>0;i--)
			{
				if(sum<money[i-1]||!book[i])
				continue;
				n=min(sum/money[i-1],book[i]);
				sum-=money[i-1]*n;
				Manum+=n;
			}
			printf("%d %d\n",Minum,num-Manum);
		}
	}
	return 0;
}