Poj 3181 (dp)

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本文介绍了一道经典的完全背包问题——DollarDays购物组合算法题。农民John在商店用有限的资金购买不同价格的工具，求解所有可能的购买组合。使用了高精度BigInteger来处理大数据情况。

Dollar Dayz

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2912		Accepted: 1158

Description

Farmer John goes to Dollar Days at The Cow Store and discovers an unlimited number of tools on sale. During his first visit, the tools are selling variously for $1, $2, and $3. Farmer John has exactly $5 to spend. He can buy 5 tools at $1 each or 1 tool at $3 and an additional 1 tool at $2. Of course, there are other combinations for a total of 5 different ways FJ can spend all his money on tools. Here they are:

        1 @ US$3 + 1 @ US$2

        1 @ US$3 + 2 @ US$1

        1 @ US$2 + 3 @ US$1

        2 @ US$2 + 1 @ US$1

        5 @ US$1

Write a program than will compute the number of ways FJ can spend N dollars (1 <= N <= 1000) at The Cow Store for tools on sale with a cost of $1..$K (1 <= K <= 100).

Input

A single line with two space-separated integers: N and K.

Output

A single line with a single integer that is the number of unique ways FJ can spend his money.

Sample Input

5 3

Sample Output

Source

USACO 2006 January Silver

dp入门题。看成完全背包即可，要用高精度。

import java.math.BigInteger;
import java.util.Scanner;

public class Main {

	static final int maxn = 1000 + 5;
	static final int maxk = 100 + 5;
	static BigInteger dp[][]=new BigInteger[maxk][maxn];

	static void solve() {
		for (int i = 0; i < maxk; i++) dp[i][0] = BigInteger.ONE;
		for (int i = 1; i < maxn; i++) dp[0][i] = BigInteger.ZERO;
		for (int i = 1; i < maxk; i++) {
			for (int j = 1; j < maxn; j++) {
				if (j >= i) dp[i][j] = dp[i - 1][j].add(dp[i][j - i]);
				else dp[i][j] = dp[i - 1][j];
			}
		}
	}

	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		int n, k;
		solve();
		while (cin.hasNext()) {
			n = cin.nextInt();
			k = cin.nextInt();
			System.out.println(dp[k][n]);
		}
	}

}