HDU 5527 Too Rich (好题 贪心 DFS)

Too Rich

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

Total Submission(s): 228    Accepted Submission(s): 69

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

317 8 4 2 0 0 0 0 0 0 0100 99 0 0 0 0 0 0 0 0 02015 9 8 7 6 5 4 3 2 1 0

 

Sample Output

9-136

 

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

 

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

题目大意：有1，5，10，20，50，100，200，500，1000，2000十种面值的钱币各ci个，现在要凑出p元，问最多可以用多少钱币凑出

题目分析：长春的金牌题。。。首先按贪心的思想，用尽可能多的小面值钱币，前提是小面值钱币可以凑出当前需要的钱数，所以从大面值的开始决策，比如现在到第idx个面值的钱币，要凑x元，又用1~idx-1的钱币可以凑出的总金额为y元，那么当前面值我需要用(x - y) / c[i]个，当然如果这个值小于0，就不用当前面值的钱币，注意如果c[i]不能整除(x- y)，则需要多用一个idx的钱币，因为剩下的钱不够，比如有 10 20 20 50 50，现在要凑110，110 - 10 - 20 - 20 = 60，50不能整除60，则就需要两个50的，因为只用一个50的话，剩下的凑不出60，还有一点要注意的是20不能整除50，200不能整除500，因此我们算个数的时候有时需要多加一个，比如20 20 20 50，现在要凑50，因为剩下3个20，一共可以凑60，按照贪心策略那个单独的50就不会被选了，因此这里需要强制选一个50，200和500同理

#include <cstdio>#include <cstring>#include <algorithm>#define ll long longusing namespace std;int p, ans, c[11];int val[11] = {0, 1, 5, 10, 20, 50, 100, 200, 500, 1000, 2000}; ll sum[11];void DFS(int rest, int idx, int cnt){    if(rest < 0)        return;    if(idx == 0)    {        if(rest == 0)            ans = max(ans, cnt);        return;    }    ll cur = max(rest - sum[idx - 1], 0ll);    int curnum = cur / val[idx];    if(cur % val[idx])        curnum ++;    if(curnum <= c[idx])        DFS(rest - curnum * val[idx], idx - 1, cnt + curnum);    curnum ++;    if(curnum <= c[idx])        DFS(rest - curnum * val[idx], idx - 1, cnt + curnum);}int main(){    int T;    scanf("%d", &T);    while(T --)    {        memset(sum, 0, sizeof(sum));        ans = -1;        scanf("%d", &p);        for(int i = 1; i <= 10; i++)            scanf("%d", &c[i]);        for(int i = 1; i <= 10; i++)            sum[i] = sum[i - 1] + (ll)(val[i] * c[i]);        DFS(p, 10, 0);        printf("%d\n", ans);    }}