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poj3181 Dollar Dayz (DP+大数)

来源:程序员人生   发布时间:2016-06-01 17:18:11 阅读次数:898次

Dollar Dayz
Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 3181
Appoint description:

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

5
1个DP题 不过由于数太大了。。。所以我选用的java...

其中dp数组存贮组成当前数的个数

import java.math.*; import java.util.*; public class Main { public static void main(String[] args) { Scanner sca =new Scanner(System.in); int k,n; while(sca.hasNext()) { n=sca.nextInt(); k=sca.nextInt(); BigInteger dp []=new BigInteger[1005]; for(int i=0;i<1005;i++) dp[i]=BigInteger.ZERO; dp[0]=BigInteger.ONE; for(int i=1;i<=k;i++){ for(int j=0;j<=n;j++){ if(j>=i&&dp[j-i]!=BigInteger.ZERO){ dp[j]=dp[j].add(dp[j-i]); } } } System.out.println(dp[n]); } } }



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