Uva--147（动规）

2014-07-30 12:05:23

 Dollars 

New Zealand currency consists of $100, $50, $20, $10, and $5 notes and $2, $1, 50c, 20c, 10c and 5c coins. Write a program that will determine, for any given amount, in how many ways that amount may be made up. Changing the order of listing does not increase the count. Thus 20c may be made up in 4 ways: 1  20c, 2  10c, 10c+2  5c, and 4  5c.

Input

Input will consist of a series of real numbers no greater than $300.00 each on a separate line. Each amount will be valid, that is will be a multiple of 5c. The file will be terminated by a line containing zero (0.00).

Output

Output will consist of a line for each of the amounts in the input, each line consisting of the amount of money (with two decimal places and right justified in a field of width 6), followed by the number of ways in which that amount may be made up, right justified in a field of width 17.

Sample input

0.20
2.00
0.00

 

Sample output

  0.20                4
  2.00              293

思路：经典的完全背包。（1）定义币值 （2）dp【0】 = 1 （给好种子） （3）从前往后推的dp

/*************************************************************************

    > File Name: j.cpp
    > Author: Nature
    > Mail: 564374850@qq.com 
    > Created Time: Tue 29 Jul 2014 07:49:33 PM CST
************************************************************************/

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long

int main(){
    ll v[11] = {10000,5000,2000,1000,500,200,100,50,20,10,5};
    ll dp[30005] = {1};
    for(int i = 0; i < 11; ++i)
        for(int j = 0; j + v[i] <= 30000; ++j)
            dp[j + v[i]] += dp[j];
    int n,m;
    while(scanf("%d.%d",&n,&m) == 2 && (n || m)){
        printf("%3d.%02d%17lld\n",n,m,dp[n * 100 + m]);
    }
    return 0;
}