POJ 1252 Euro Efficiency【完全背包正负】

Euro Efficiency

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3649		Accepted: 1561

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

Source

Northwestern Europe 2002

嗯，题目大意就是说，给出6种面额的现金，问从1到100用这些面额的纸币表示出来用最少数量的纸币数，然后求的是从1到100平均所用的纸币数以及1到100之间用纸币数最多的那个数所用的纸币数，这里一个数既可以相加得到也可以相减得到，先相加再相减，但是反过来好像也可以，我还没理解透彻。。。这里要求的是尽量少的纸币数，f数组记录所用的纸币数，所以每次所加的价值就变为1了，首先将f数组初始化一个较大的数，这个是相当于背包要恰好装满的，然后f[0]=0（最后就是输出的时候，double型的输入是%lf 而输出是%f 若用C++提交%lf 输出也可以）

#include <iostream>
#include<cstdio>
#include<cstring>
#define maxn 5000+10
using namespace std;
int f[maxn],p[7];
int min(int a,int b)
{
    return a<b?a:b;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        for(int i=0;i<6;++i)
            scanf("%d",&p[i]);
        memset(f,0,sizeof(f));
        for(int i=1;i<maxn;++i)
            f[i]=9999;
        int cnt=0,rec=-1;
        for(int i=0;i<6;++i)
        {
            for(int j=p[i];j<maxn;++j)
                f[j]=min(f[j],f[j-p[i]]+1);
        }
        for(int i=0;i<6;++i)
        {
            for(int j=maxn-p[i];j>=0;j--)
                f[j]=min(f[j],f[j+p[i]]+1);
        }
        for(int i=0;i<=100;++i)
        {
            cnt+=f[i];
            rec=rec>f[i]?rec:f[i];
        }
        printf("%.2f %d\n",cnt*1.0/100,rec);
    }
    return 0;
}