Too Rich（贪心+DFS）

Posted 2021-01-10 fighting-sh

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

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <string>
 4 #include <cmath>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <queue>
 8 #include <stack>
 9 #include <vector>
10 #include <set>
11 #include <map>
12 #define maxn 50010
13 #define lson l,mid,rt<<1
14 #define rson mid+1,r,rt<<1|1
15 const long long INF=0x3f3f3f3f3f3f3f3f;
16 using namespace std;
17 long long coin[] = {0,1, 5, 10, 20, 50, 100, 200, 500, 1000, 2000};
18 long long a[12];
19 long long ans;
20 
21 void dfs(int pos,long long sum,long long num){
22     if(sum==0){
23         ans=min(ans,num);
24         return;
25     }
26     if(pos<1){
27         return;
28     }
29     long long tmp=min(a[pos],sum/coin[pos]);
30     dfs(pos-1,sum-tmp*coin[pos],num+tmp);
31     if(tmp>0){//去掉一个硬币的情况
32         tmp--;
33         dfs(pos-1,sum-tmp*coin[pos],num+tmp);
34     }
35 }
36 
37 int main(){
38     int t;
39     scanf("%d",&t);
40     while(t--){
41         int total=0;
42         long long sum=0;
43         for(int i=0;i<11;i++){
44             scanf("%lld",&a[i]);
45             sum+=coin[i]*a[i];
46             total+=a[i];
47         }
48         total-=a[0];//a[0]表示p
49         if(sum<a[0]){
50             puts("-1");
51             continue;
52         }
53         sum-=a[0];
54         ans=INF;
55         dfs(10,sum,0);
56         if(ans==INF){
57             puts("-1");
58         }
59         else{
60             printf("%lld
",total-ans);
61         }
62     }
63 }

View Code