POJ Euro Efficiency 1252

Euro Efficiency

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4109		Accepted: 1754

Description

On January 1st 2002, The Netherlands, and several other European countries abandoned their national currency in favour of the Euro. This changed the ease of paying, and not just internationally.
A student buying a 68 guilder book before January 1st could pay for
the book with one 50 guilder banknote and two 10 guilder banknotes,
receiving two guilders in change. In short:50+10+10-1-1=68. Other ways
of paying were: 50+25-5-1-1, or 100-25-5-1-1.Either way, there are
always 5 units (banknotes or coins) involved in the payment process, and
it

could not be done with less than 5 units.

Buying a 68 Euro book is easier these days: 50+20-2 = 68, so only 3
units are involved.This is no coincidence; in many other cases paying
with euros is more efficient than paying with guilders. On average the
Euro is more efficient. This has nothing to do, of course, with the
value of the Euro, but with the units chosen. The units for guilders
used to be: 1, 2.5, 5, 10, 25, 50,whereas the units for the Euro are: 1,
2, 5, 10, 20, 50.

For this problem we restrict ourselves to amounts up to 100 cents.
The Euro has coins with values 1, 2, 5, 10, 20, 50 eurocents. In paying
an arbitrary amount in the range [1, 100] eurocents, on average 2.96
coins are involved, either as payment or as change. The Euro series is
not optimal in this sense. With coins 1, 24, 34, 39, 46, 50 an amount of
68 cents can be paid using two coins.The average number of coins
involved in paying an amount in the range [1, 100] is 2.52.

Calculations with the latter series are more complex, however. That
is, mental calculations.These calculations could easily be programmed in
any mobile phone, which nearly everybody carries around nowadays.
Preparing for the future, a committee of the European Central Bank is
studying the efficiency of series of coins, to find the most efficient
series for amounts up to 100 eurocents. They need your help.

Write a program that, given a series of coins, calculates the
average and maximum number of coins needed to pay any amount up to and
including 100 cents. You may assume that both parties involved have
sufficient numbers of any coin at their disposal.

Input

The
first line of the input contains the number of test cases. Each test
case is described by 6 different positive integers on a single line: the
values of the coins, in ascending order. The first number is always 1.
The last number is less than 100.

Output

For
each test case the output is a single line containing first the average
and then the maximum number of coins involved in paying an amount in the
range [1, 100]. These values are separated by a space. As in the
example, the average should always contain two digits behind the decimal
point. The maximum is always an integer.

Sample Input

3
1 2 5 10 20 50
1 24 34 39 46 50
1 2 3 7 19 72

Sample Output

2.96 5
2.52 3
2.80 4
完全背包问题，钞票的和包括加和减，因此输入1到6个正数，可将其相反数存到7到12中
PS：G++下double输出为%.2f,C++下double输出为%.2lf

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int dp[];
int a[],t;
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        memset(a,,sizeof(a));
        double ans=0.0;
        int pos=-;
        for(int i=;i<=;i++)
        {
            scanf("%d",&a[i]);
            a[i+]=-a[i];
        }
        for(int i=;i<=;i++)
        {
            dp[i]=INF;
        }
        dp[]=;
        for(int i=;i<=;i++)
        {
            if(a[i]>)
            {
                for(int j=a[i];j<=;j++)
                {
                    dp[j]=min(dp[j],dp[j-a[i]]+);
                }
            }
            else
            {
                for(int j=;j>=-a[i];j--)
                {
                    dp[j+a[i]]=min(dp[j+a[i]],dp[j]+);
                }
            }
        }
        for(int i=;i<=;i++)
        {
            ans+=dp[i];
            pos=max(pos,dp[i]);
        }
        printf("%.2f %d\n",ans/(100.0),pos);
    }
    return ;
}