Problem f

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
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    根据题意，即给出一个数，然后给出5个数，分别是1毛5毛10毛50毛100毛，来凑出前面的数，输出最少用几张，最多用几张。
   
   
   
   
   
   
   
   

    
    
    
    此题，最少好做，就用贪心算法就可，而最多就稍微麻烦一点了，请看代码注释。
   
   
   
   
   
   
   
   
 
   
   
   
   
   
   
   
   

    
    
    
    思路就是， 求最少时，就用贪心算法，很简单。当求最多时，转化一下，就是求剩下钱的最少值：第一组例子，33 6 6 6 6 6，一共是996毛减去33毛，就是剩下的963毛，下面的问题 就是 963毛用现有的钱数最少用几张凑出。这样33毛不就是最对了么，反正钱的总张数是一定的。
   
   
   
   
   
   
   
   

    
    
    
    #include<iostream> using namespace std; int f1(int m,int a1,int a5,int a10,int a50,int a100) {     int mm,m2=0,n,sum=0;     m2=a1+a5+a10+a50+a100;     mm=a1+5*a5+10*a10+50*a50+100*a100;     n=mm-m;     for(;n >= 100&&a100 > 0;--a100)     {          int m1=0;           ++m1;         n = n-100;         sum=sum+m1;     }     for(;n >= 50&&a50 > 0;--a50)     {           int m1=0;           ++m1;         n = n-50;         sum=sum+m1;     }     for(;n >= 10&&a10 > 0;--a10)     {           int m1=0;           ++m1;         n = n-10;         sum=sum+m1;     }     for(;n >= 5&&a5 > 0;--a5)     {          int m1=0;           ++m1;         n = n-5;         sum=sum+m1;     }     for(;n >= 1&&a1 > 0;--a1)     {          int m1=0;         n = n-1;         ++m1;          sum=sum+m1;     }     if(n == 0)         return (m2-sum);     else         return -1; } int f2(int n,int a1,int a5,int a10,int a50,int a100) {     int sum=0;    for(;n >= 100&&a100 > 0;--a100)     {          int m1=0;           ++m1;         n = n-100;         sum=sum+m1;     }     for(;n >= 50&&a50 > 0;--a50)     {         int m1=0;           ++m1;         n = n-50;         sum=sum+m1;     }     for(;n >= 10&&a10 > 0;--a10)     {         int m1=0;           ++m1;         n = n-10;         sum=sum+m1;     }     for(;n >= 5&&a5 > 0;--a5)     {          int m1=0;           ++m1;         n = n-5;         sum=sum+m1;     }     for(;n >= 1&&a1 > 0;--a1)     {          int m1=0;         n = n-1;         ++m1;         sum=sum+m1;     }     if(n == 0)         return sum;     else         return -1; } int main() {     int m,k,Min,Max;     int n,m1=0,m2,a1,a5,a10,a50,a100;     cin>>k;     while(k)     {         --k;     cin>>n>>a1>>a5>>a10>>a50>>a100;     Min = f2(n,a1,a5,a10,a50,a100);     Max = f1(n,a1,a5,a10,a50,a100);     cout<<Min<<" "<<Max<<endl;     } }