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[] = {0,1, 5, 10, 20, 50, 100, 200, 500, 1000, 2000};
long long a[12];
long long ans;

void dfs(int pos,long long sum,long long num){
    if(sum==0){
        ans=min(ans,num);
        return;
    }
    if(pos<1){
        return;
    }
    long long tmp=min(a[pos],sum/coin[pos]);
    dfs(pos-1,sum-tmp*coin[pos],num+tmp);
    if(tmp>0){//去掉一个硬币的情况
        tmp--;
        dfs(pos-1,sum-tmp*coin[pos],num+tmp);
    }
}

int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int total=0;
        long long sum=0;
        for(int i=0;i<11;i++){
            scanf("%lld",&a[i]);
            sum+=coin[i]*a[i];
            total+=a[i];
        }
        total-=a[0];//a[0]表示p
        if(sum<a[0]){
            puts("-1");
            continue;
        }
        sum-=a[0];
        ans=INF;
        dfs(10,sum,0);
        if(ans==INF){
            puts("-1");
        }
        else{
            printf("%lld
",total-ans);
        }
    }
}