UVA - 147 - Dollars （集合上的动态规划）

UVA - 147

Dollars

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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

Source

Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 3. Problem Solving Paradigms :: Dynamic Programming :: Coin
Change - Classical

Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 5. Dynamic Programming

Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Problem Solving Paradigms :: Dynamic Programming :: Coin
Change (CC)

Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Problem Solving Paradigms :: Dynamic Programming :: Coin
Change (CC)

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思路：简单的动态规划问题，因为每个面值都是5cent的整数倍，所以将所有面值都转化为5cent的倍数，即{1, 2, 4, 10, 20, 40, 100, 200, 400, 1000, 2000};

输入的那个数也转化为5的倍数，注意浮点数问题

然后dp预处理下，输出记得域宽要按照要求来，这里dp因为会爆int，要开到LL型

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;

const double eps = 1e-7;
double n;
LL dp[6005];
int val[11] = {1, 2, 4, 10, 20, 40, 100, 200, 400, 1000, 2000};

void init() {
	memset(dp, 0, sizeof(dp));
	dp[0] = 1;
	for(int i = 0; i < 11; i++) {
		for(int j = 0; j < 6005; j++) {
			if(j >= val[i]) dp[j] += dp[j-val[i]];
		}
	}
}

int main() {
	init();
	while(scanf("%lf", &n) != EOF) {
		if( (n - 0) < eps) break;
		int num = (int)( (n + 0.01) / 0.05 );
		printf("%6.2lf%17lld\n", n, dp[num]);
	}
	return 0;
}