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
题意：经典问题，换钱。
精度呀马丹。打表速度可能快点。。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
typedef unsigned long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
int a[]={5,10,20,50,100,200,500,1000,2000,5000,10000};
LL dp[300100];
int n;
int main()
{
double x;
while(~scanf("%lf",&x)&&x!=0)
{
n=(int)(x*100+0.5);//+0.5注意精度
// cout<<"23333 "<<n<<endl;
memset(dp,0,sizeof(dp));
dp[0]=1;
REP(i,11)//对于每一种货币考虑加入到j中
{
REPF(j,a[i],n)
dp[j]+=dp[j-a[i]];
}
printf("%6.2f%17lld\n",x,dp[n]);
}
}