1 005 天才的硬币

"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

 这个题！！最多的可以用手里剩的最小的来求 不像某雨 求了那么多哈

#include<iostream>
#include<string.h>
#include<set>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<numeric>
#include<math.h>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<string>
using namespace std;
int b[10]={0,1,5,10,50,100};
int main()
{
    int t;
    int p,r;
    int a[10],c[10],e[10];
    int i,j,k,sum;

    cin>>t;;
    while(t--)
    {
        sum=0;
        cin>>p;            // 输入价钱数
        r=p;
        for(i=1;i<=5;i++)
        {
            cin>>a[i];
            sum=sum+b[i]*a[i];    //判断价钱总数
        }
        for(i=5;i>0;i--)
        {
            if(r/b[i]<a[i])
            {
                c[i]=r/b[i];
                r=r-b[i]*c[i];    //如果 总数/价钱数 < 这个价钱的个数 那么 从c【i]赋值；  r用来保留剩下的；

            }
            else
            {
                c[i]=a[i];              //如果没办法减去 那么直接加上价钱的个数
                r=r-c[i]*b[i];
            }
        }
        if(r!=0)
        {
            cout<<"-1"<<" "<<"-1"<<endl;               //如果r最后没有被剪完就说明！1！他跪了！！！！他不行了啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊我的逻辑！！

        }else
        {
             k=sum-p;                       //这一步我十分佩服我自己相当的机智 一个机智的boy  如果要硬币最多 手里的不就最少吗！！一减不就行了吗！！！
            for(i=5;i>0;i--)
            {
                if(k/b[i]<a[i])
                {
                    e[i]=k/b[i];
                    k=k-b[i]*e[i];
                }
                else
                {
                    e[i]=a[i];
                    k=k-e[i]*b[i];
                }
            }
            if(k==0)
            {
                cout<<c[1]+c[2]+c[3]+c[4]+c[5] ;cout<<" " ;cout<<a[1]+a[2]+a[3]+a[4]+a[5]-e[1]-e[2]-e[3]-e[4]-e[5]<<endl;
            }

        }

    }
}