Too Rich

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

Total Submission(s): 1666    Accepted Submission(s): 422

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

 

听说是一个金牌题，确实很巧妙。

开始的想法是，多重背包，用二进制优化成01背包，d[i] ：钱币为 i 的最多钱币数目，但是开不了这么大的数组，用map映射一下，或者改成记忆化搜索写法，但是是不可能的，记忆化还是得有一个vis数组~~~。

 

 

正解：

先转换思路，凑P求最多硬币，转为求mon - P 的最少硬币数。

然后贪心，但是会像记忆化搜索一样要回溯。直接贪心不太好。但是，硬币问题有一个特点，如果所有的钱币如： 1  ,   5   ,   10  ,   20  , 40   ,   80  ,   800，你已经看出来了，就是这种前面是后面的因子，这样就不可能说要从后往前的过程中要回溯。直接从后往前贪心。

 

怎么转为这种情况呢，发现只要改变两个数  50,500，改成 100,1000来用，就可以了。

#include 
       
        using namespace std;const int inf = 0x3f3f3f3f;int c[30];int v[10] = {
     1,5,10,20,50,100,200,500,1000,2000};int num[10];int sum,mon,p,ans;int solve(int x) {    int ret = 0,need;    for(int i = 9; i >=0; i--) {        if(v[i]==500) {            need = x/1000;            need = min(need,num[i]/2);            ret +=2*need;            x-=need*1000;        }        else if(v[i]==50) {            need = x/100;            need = min(need,num[i]/2);            ret +=2*need;            x-=need*100;        }        else {            need = x/v[i];            need = min(need,num[i]);            ret +=need;            x -=need*v[i];        }    }    if(x) return inf;    return ret;}int main(){    //freopen("in.txt","r",stdin);    int T;    scanf("%d",&T);    while(T--) {        sum = mon = 0;        scanf("%d",&p);        for(int i = 0; i < 10; i++) {            scanf("%d",&num[i]);            sum +=num[i];            mon +=num[i]*v[i];        }        p = mon - p;    //最少的硬币去兑换        if(p<0) {            puts("-1");            continue;        }        ans = inf;        ans = min(ans,solve(p));        if(num[7]) {            p-=500;num[7]--;            if(p>=0)                ans = min(ans,solve(p)+1);            p+=500;num[7]++;        }        if(num[4]) {            p-=50;num[4]--;            if(p>=0)                ans = min(ans,solve(p)+1);            p+=50;            num[4]++;        }        if(num[4]&&num[7]) {            p-=550;num[4]--;num[7]--;            if(p>=0)                ans = min(ans,solve(p)+2);        }        if(ans==inf) puts("-1");        else printf("%d\n",sum-ans);    }    return 0;}