UVA147——DP（找钱2）——Dollars

Description

 

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
大意：同UVA647 double转换成int为什么要加0.5.....

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long  dp[30005][15];
int v[11] = {5,10,20,50,100,200,500,1000,2000,5000,10000};
long long  DP(long long  i,long long  j)
{
    if(dp[i][j] != -1)
        return dp[i][j];
    dp[i][j] = 0;
    for(int k = 0 ; i-k*v[j] >= 0; k++)
        dp[i][j] += DP(i-k*v[j],j-1);
    return dp[i][j];
}
int main()
{
    memset(dp,-1,sizeof(dp));
    double n;
    while(~scanf("%lf",&n)){
        if(n == 0.00)
            break ;
      int  m =(int)(n*100+0.5);
        for(long long  i = 0 ; i <= m; i++)
            dp[i][0] = 1;
            printf("%6.2f%17lld\n",n,DP(m,10));
    }
    return 0;
}