[題目連結](http://acm.hdu.edu.cn/showproblem.php?pid=3348)                  

coins

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2074 Accepted Submission(s): 669

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(紙幣), 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

這道題本來很簡單，求最多紙幣如果直接求那是比較費勁的，如果反過來想問題的話那就簡單很多了，具體看程式碼

#include <algorithm>
#include <string.h>
#include <iostream>
#include <stdio.h>
#include <string>
#include <vector>
 

#include <queue>
#include <map>
#include <set>
#include<cmath>
#include <iomanip>
using namespace std;
const int N = 1e6;
const double esp=1e-5;
int money[]= {1,5,10,50,100};
int Count[5];
int t,p;
int SolveMin(int p)
{
    int ans=0,m;
    for(int i=4; i>=0; i--)
    {
        m=min(p/money[i],Count[i]);
        p-=m*money[i];
        ans+=m;
    }
    if(p==0)  return ans;
    return -1;
}
int main()
{
    cin>>t;
    while(t--)
    {
        cin>>p;
        int sum=0,sumCout=0;
        for(int i=0; i<=4; i++)
        {
            cin>>Count[i];
            sum+=Count[i]*money[i]; sumCout+=Count[i];
        }
        if(sum<p) printf("-1 -1\n");
        else
        {
            int s=SolveMin(p);
            cout<<s<<" "<<sumCout-SolveMin((sum-p))<<endl;
        }
    }
    return 0;
}