UVA 147 - Dollars(完全背包)

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

                                                 

题意类似于上一道UVA674

却又无数坑点！答案应使用long long，double 数转化为int 应该注意精度问题 int nn = (int) (n * 20 + 0.5); UVA 是 %lld 输出的~

CODE;

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
const int inf=0xfffffff;
typedef long long ll;
using namespace std;

const int maxn = 30010/5;
int a[11] = {1,2,4,10,20,40,100,200,400,1000,2000};
ll dp[maxn];
int main()
{
double n;

while(~scanf("%lf", &n)){
if(n == 0.0) break;
int nn = (int)(n *20 + 0.5);
memset(dp, 0, sizeof(dp));
dp[0] = 1;
for(int i = 0; i < 11; ++i){
for(int j = a[i]; j <=nn; ++j){
dp[j] = dp[j] + dp[j - a[i]];
}
}
printf("%6.2lf%17lld\n", n, dp[nn]);
}
return 0;
}