Too Rich（贪心+DFS）

Too Rich

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

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2032    Accepted Submission(s): 529

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 pdollars 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 $10 coins, four $5 coins, and eight $1 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-重现赛（感谢东北师大）

 

这题求的是用最多的硬币凑成P元，我们可以反着想，用最少的硬币凑成剩下的钱，这样就转换成一道经典的贪心问题

但是要注意的是，大的硬币不一定是小的硬币的整数倍，所以需要用DFS搜索出最优解，具体看代码

 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <cmath>
 #include <iostream>
 #include <algorithm>
 #include <queue>
 #include <stack>
 #include <vector>
 #include <set>
 #include <map>
 #define maxn 50010
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 const long long INF=0x3f3f3f3f3f3f3f3f;
 using namespace std;
 long long coin[] = {,, , , , , , , , , };
 long long a[];
 long long ans;

 void dfs(int pos,long long sum,long long num){
     if(sum==){
         ans=min(ans,num);
         return;
     }
     if(pos<){
         return;
     }
     long long tmp=min(a[pos],sum/coin[pos]);
     dfs(pos-,sum-tmp*coin[pos],num+tmp);
     if(tmp>){//去掉一个硬币的情况
         tmp--;
         dfs(pos-,sum-tmp*coin[pos],num+tmp);
     }
 }

 int main(){
     int t;
     scanf("%d",&t);
     while(t--){
         int total=;
         long long sum=;
         for(int i=;i<;i++){
             scanf("%lld",&a[i]);
             sum+=coin[i]*a[i];
             total+=a[i];
         }
         total-=a[];//a[0]表示p
         if(sum<a[]){
             puts("-1");
             continue;
         }
         sum-=a[];
         ans=INF;
         dfs(,sum,);
         if(ans==INF){
             puts("-1");
         }
         else{
             printf("%lld\n",total-ans);
         }
     }
 }