uva 147 Dollars (dp + 完全背包)

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

0.202.000.00

Sample output

  0.20                4  2.00              293

题意：你有 $100, $50, $20, $10, and $5 notes and $2, $1, 50c, 20c, 10c and 5c的钱币，现在给你一个总价值，你要求出用这些钱币能组合出该价值的数量。

思路：完全背包。状态转移方程为d[j] = d[j] + d[j - coin[i]]。水水的。不过要注意double转int的精度误差和数组要开成longlong。因为价值大得时候组合数量超过了2^31 - 1。

代码：

#include <stdio.h>#include <string.h>const int coin[12] = {10000, 5000, 2000, 1000, 500, 200, 100, 50, 20, 10, 5};double value;long long d[30005];int i, j;int main() {d[0] = 1;for (i = 0; i < 11; i ++)for (j = coin[i]; j <= 30000; j ++) { d[j] = d[j] + d[j - coin[i]];}while (~scanf("%lf", &value) && value) {printf("%6.2lf%17lld\n", value, d[int(value * 100 + 0.5)]);}return 0;}