Problem F

Problem Description

"Yakexi, this is the best age!" Dong MW works hard and get high pay, he has many 1 Jiao and 5 Jiao banknotes(纸币), some day he went to a bank and changes part of his money into 1 Yuan, 5 Yuan, 10 Yuan.(1 Yuan = 10 Jiao)
"Thanks to the best age, I can buy many things!" Now Dong MW has a book to buy, it costs P Jiao. He wonders how many banknotes at least,and how many banknotes at most he can use to buy this nice book. Dong MW is a bit strange, he doesn't like to get the change, that is, he will give the bookseller exactly P Jiao.

 

Input

T(T<=100) in the first line, indicating the case number. T lines with 6 integers each: P a1 a5 a10 a50 a100 ai means number of i-Jiao banknotes. All integers are smaller than 1000000.

 

Output

Two integers A,B for each case, A is the fewest number of banknotes to buy the book exactly, and B is the largest number to buy exactly.If Dong MW can't buy the book with no change, output "-1 -1".

 

Sample Input

   
   
   
   

    
    
    
    3
33 6 6 6 6 6
10 10 10 10 10 10
11 0 1 20 20 20
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    6 9
1 10
-1 -1
   
   
   
   

 简单题意：

  Dong拥有很多1角、5角，并且换到了1元、5元、10元的纸币。他要买一本书，并且他不喜欢找零钱，所以编写一个程序求出最小的纸币张数和最大的纸币张数。

解题思路形成过程：

  这是考验思路以及逻辑思维能力的题目。我真是没想出来，用自己的思路编写得到的代码太长，自己都没把握AC。所以，我搜索了类似的算法，最终从无到有把这类题型烂熟于心。

感想：

  一步一步分解，逐步就能求出最佳的解。

AC代码：

 

#include <iostream>
#include <algorithm>
using namespace std;
int min(int a[],int p,int num[])
{
    int cnt=0;
    for(int i=5;i>1;i--)
    if(p>=num[i]*a[i])
    {
        cnt+=num[i];
        p-=a[i]*num[i];
    }//当书的价格比当前的钱数还要大时，执行；
    else
    {
        cnt+=p/a[i];//不能用掉所有a[i]角纸币时，只需要最高位数字就可以。这必然是最优的方式；
        p=p%a[i];
    }
    if(p>num[1])return -1;
    else return cnt+p;
}
int max(int a[],int num[],int sum[],int p)
{
    int cnt=0;
    for(int i=5;i>1;i--)
    {
        if(p<=sum[i-1])continue;
        else
        {
            int b,c,d;
            b=(p-sum[i-1])/a[i];
            cnt+=b;
            c=(p-sum[i-1])%a[i];
            if(c!=0)
            {
                cnt+=1;
                d=b+1;
            }
        else d=b;
            p-=a[i]*d;
        }

    }
    if(p>num[1])return -1;
    else return cnt+p;
}
void greedy_select(int a[],int num[],int p)
{
    int sum[6];
    sum[0]=0;
    sum[1]=num[1]*1;
    for(int i=2;i<5;i++)
        sum[i]=sum[i-1]+a[i]*num[i];
    int minNum,maxNum;
    minNum=min(a,p,num);
    if(minNum==-1)cout<<"-1 -1"<<endl;
    else
    {
         maxNum=max(a,num,sum,p);
        if(maxNum==-1)cout<<"-1 -1"<<endl;
        else  cout<<minNum<<" "<<maxNum<<endl;
    }
}
int main()
{
    int n;
    while(cin>>n)
    {
        for(int i=0;i<n;i++)
        {
            int num[6],p,a[6]={0,1,5,10,50,100},sum=0;
            cin>>p;
            for(int j=1;j<6;j++)
            {
                cin>>num[j];
                sum+=num[j]*a[j];
            }
            if(sum<p)cout<<"-1 -1"<<endl;
            else greedy_select(a,num,p);
        }
    }
    return 0;
}