【hdu 5527】 [2015ACM/ICPC亚洲区长春站] Too Rich　贪心

Too Rich

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 673 Accepted Submission(s): 173

Problem Description

You are a rich person, and you think your wallet is too heavy and full now. So you want to give me some money by buying a lovely pusheen sticker which costs p dollars from me. To make your wallet lighter, you decide to pay exactly p dollars by as many coins and/or banknotes as possible.

For example, if p=17 and you have two 10coins,four10 coins, four 5 coins, and eight 1coins,youwillpayitbytwo1 coins, you will pay it by two 5 coins and seven $1 coins. But this task is incredibly hard since you are too rich and the sticker is too expensive and pusheen is too lovely, please write a program to calculate the best solution.

Input

The first line contains an integer T indicating the total number of test cases. Each test case is a line with 11 integers p,c1,c5,c10,c20,c50,c100,c200,c500,c1000,c2000, specifying the price of the pusheen sticker, and the number of coins and banknotes in each denomination. The number ci means how many coins/banknotes in denominations of i dollars in your wallet.

1≤T≤20000

0≤p≤109

0≤ci≤100000

Output

For each test case, please output the maximum number of coins and/or banknotes he can pay for exactly p dollars in a line. If you cannot pay for exactly p dollars, please simply output ‘-1’.

Sample Input

3

17 8 4 2 0 0 0 0 0 0 0

100 99 0 0 0 0 0 0 0 0 0

2015 9 8 7 6 5 4 3 2 1 0

Sample Output

9

-1

36

Source

2015ACM/ICPC亚洲区长春站

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5527

题意：

选出凑成Ｎ的最多的钱数；

思路：

从大面值到小面值贪心，剩下的只要可以被取到便取最少的大面值的；因为５００，２００等非倍数的存在，

反例：１５０，用５０，２０取，故取时500,50，等可能多取一个即可；

代码：

#include<iostream>
#include<stdio.h>
using namespace std;
int v[11]={0,1,5,10,20,50,100,200,500,1000,2000};
int s[11];
int ans;int N;
long long T[11];
void dfs(int p,int n,int d)
{

if(p<0) return ;
if(d==0)
{
if(!p)
ans=max(ans,n);
return ;
}
long long tmp=max((long long)0,p-T[d-1]);
int k=tmp/v[d]+(tmp%v[d]?1:0);

if(k>s[d]) return ;
dfs(p-k*v[d],k+n,d-1);

k++;
if(k<=s[d])
dfs(p-k*v[d],k+n,d-1);
}

int main()
{
int c;
scanf("%d",&c);
while(c--)
{
scanf("%d",&N);
for(int i=1;i<=10;i++)
scanf("%d",&s[i]);
T[0]=0;
for(int i=1;i<=10;i++)
{
T[i]=T[i-1]+(long long)s[i]*(long long)v[i];
}
ans=0;
dfs(N,0,10);

if(!ans) printf("-1\n");
else printf("%d\n",ans);
}

}