Too Rich（贪心加搜索）

个人心得：10月份月赛题目，很low，就过了一道水题而且是把所有猜测都提交才过的。这段时间不知道忙什么去了，

也没怎么刷题感觉自己越来越差，还不如新来的大一学弟呢，别人起码天天刷代码到半夜，比起刚在区域赛拿银的学长们，

溜了溜了。慢慢磨练吧。

题目心得：这题意思很简单，就是用最多的硬币去负款，我比赛时是从正面开始的，发现自己low一是这样麻烦而且漏洞多，

后面看了题解他是从后面开始的。

要用尽量多的硬币凑P元钱，假设我现在硬币总共的面值和为M，那么换个思路，我现在要从总共的硬币中拿走尽量少的硬币，使剩下的硬币为P元。

那么问题转化成了，用尽量少的硬币去凑出M−P元，这个题目就有点熟悉了，不过这里有两个特殊的地方，一个是每种硬币都有数量限制，一个是面值大的硬币并不是面值小的硬币的整数倍。

假设没有数量限制，那么就是简单的贪心，尽量用面值大的硬币去凑出M−P元就好了。但是这个题用这个思路的话，会有这种情况。

假设现在有20,20,20,50这四个硬币，要去凑60元，如果以之前的思路贪心的话，那么是无解的，但是这里可以不用50元这个硬币，而用3个20元的去凑出60元。

那么对于这种情况，就要尝试一下，对于每种硬币，去掉一个这种硬币后的解。

题目：

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 pp dollars from me. To make your wallet lighter, you decide to pay exactly pp dollars by as many coins and/or banknotes as possible. 

For example, if p=17p=17 and you have two $10$10 coins, four $5$5 coins, and eight $1$1coins, you will pay it by two $5$5 coins and seven $1$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.

InputThe first line contains an integer TT 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,c2000p,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 cici means how many coins/banknotes in denominations of ii dollars in your wallet. 

1≤T≤200001≤T≤20000 
0≤p≤1090≤p≤109 
0≤ci≤1000000≤ci≤100000OutputFor each test case, please output the maximum number of coins and/or banknotes he can pay for exactly pp dollars in a line. If you cannot pay for exactly pp 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

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<iomanip>
#include<algorithm>
using namespace std;
const int maxn=100000005;
long long   ans=maxn;
int number[11];
int next[10]={1,5,10,20,50,100,200,500,1000,2000};
void dfs(int i,long long  sum,long long t){
    int next[10]={1,5,10,20,50,100,200,500,1000,2000};
     if(sum==0){
        ans=min(ans,t);
        return ;
     }
     if(i<0) return;
     long long  c=min((long long )number[i],sum/next[i]);
     dfs(i-1,sum-c*next[i],t+c);
     if(c>0)
     {
         c--;
         dfs(i-1,sum-c*next[i],t+c);
     }
}
int main()
{

      int t;
      cin>>t;
      int next[10]={1,5,10,20,50,100,200,500,1000,2000};
      while(t--){
               long long p,q=0;
               ans=maxn;
               int sum=0;
               cin>>p;
               for(int i=0;i<10;i++){
                 cin>>number[i];
                 q+=number[i]*next[i];
                 sum+=number[i];
               }
               long long s=q-p;
               if(s<0)
               {
                   cout<<-1<<endl;
                   continue;
               }
               dfs(9,s,0);
               if(ans==maxn)
                cout<<-1<<endl;
               else cout<<sum-ans<<endl;
               }
    return 0;
}