问题描述："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
意思：有1角，5角，10角，50角，100角的纸币，一本书x角，求花费纸币的最多数量和最小数量
思路：一种想法：因为我们要求的是花的最多数量纸币，所以就是要保证手上的纸币数量最少！！这样想的话问题就比较简单了，就转化为最少数量问题了。假设手上总共有p毛，而价格为q毛，我们用手上最少的数量的纸币去凑（p-q）毛，然后再用总数量减去该最少数量即可。
代码：
#include <iostream>
using namespace std;
int a[6]={0,1,5,10,50,100};
int main()
{
int price = 0;//定义买书所需要的钱数;
int i,j,k;
int num = 0;//买书的次数
cin >> num;
while(num--)
{
int sum = 0;
int b[6],c[6],d[6];
cin >> price;
int t = price;
for(i = 1;i <= 5;i++)
{
cin >> b[i];
sum += b[i]*a[i];
}
//求最少纸币
for(i=5;i>0;i--)
{
if(price/a[i]<b[i])
{
c[i]=price/a[i];
price=price-a[i]*c[i];
}
else
{
c[i]=b[i];
price=price-c[i]*a[i];
}
}
if(price != 0)
{
cout << "-1" << " " << "-1" << endl;
}
else
{
//求最大纸币数，反着求
k = sum - t;//卖完应该应该剩余多少钱
for(i = 5;i > 0;i--)
{
if(k/a[i] < b[i])
{
d[i] = k/a[i];
k = k - d[i] * a[i];
}
else
{
d[i] = b[i];
k = k - d[i] * a[i];
}
}
int m1 = 0,m2 = 0,m3 = 0;
for(i = 1;i <= 5;i++)
{
m1 += c[i];
m2 += d[i];
m3 += b[i];
}
if(k == 0)
cout << m1 << " " << m3-m2 << endl;
}
}
return 0;
}
参考思路：http://www.cnblogs.com/der5820/p/3907645.html