HDU 5527---Too Rich(贪心+搜索)

题目链接

 

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.

1T20000
0p109
0ci100000
 

 

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
 
题意:给了 p 表示要付的钱数,一个数列v[10],分别表示  1 ,5,10,20,50,100,200,500,1000,2000 元的钱币数量,求用尽量多的钱币刚好付清 p 元,输出钱币数。
 
思路:贪心,尽量用面值小的钱币去筹,但是很可能面值小的钱币不够,所以从大面值开始考虑。初始化一个前缀和sum[12],sum[i]表示v[1]~v[i]面值的钱币和,tmp=rest-sum[i-1],表示当前面值的钱币应该付多少,cn=tmp/v[i] ,即表示当前面值的钱币应该拿出多少张,如果tmp%v[i]!=0 ,那么cn++,因为小于v[i]的钱币无法筹出足够的钱;另外要对于P=50 钱币为 20,20,20,50 时,按照贪心策略3张20为60,所以不会取50,但是用3张20 无法筹出50元,所以必须每张面值的钱币应该多考虑一张,比如对于这样的数据:
p=1020   0 0  0 49  1   0   0   0    1     0;
 
代码如下:
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
int v[12]={0,1,5,10,20,50,100,200,500,1000,2000};
int c[12],sum[12];
int p,ans;
void dfs(int i,int rest,int count)
{
    if(rest<0) return ;
    if(i==0) {
         if(rest==0) ans=max(ans,count);
         return ;
    }
    int tmp=max(0,rest-sum[i-1]);
    int cn=tmp/v[i]+(tmp%v[i]!=0);
    if(cn<=c[i])  dfs(i-1,rest-cn*v[i],count+cn);
    cn++;
    if(cn<=c[i])  dfs(i-1,rest-cn*v[i],count+cn);
}
int main()
{
    ///cout << "Hello world!" << endl;
    int T; cin>>T;
    while(T--)
    {
        scanf("%d",&p);
        for(int i=1;i<=10;i++) scanf("%d",&c[i]);
        sum[0]=0;
        for(int i=1;i<=10;i++) sum[i]=sum[i-1]+v[i]*c[i];
        ans=-1;
        dfs(10,p,0);
        printf("%d\n",ans);
    }
    return 0;
}
///1020 0 0  0 49  1   0   0   0    1     0

 

posted @ 2017-07-18 22:01  茶飘香~  阅读(...)  评论(...编辑  收藏