Dollar Dayz POJ - 3181 动态规划完全背包

最新推荐文章于 2022-04-18 19:42:48 发布

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#Dollar Dayz POJ - 3

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本文介绍了一个购物组合计算问题：农民约翰有N美元，在一家商店里购买工具，每件工具的价格从1美元到K美元不等，目标是计算出所有可能的购物组合数量，使约翰恰好花完他的钱。

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

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <math.h>
#include <algorithm>
#define LL long long
#define INF 1000000000000000000   //I64d最大是 10^18次方
#define MAX 200010
using namespace std;
int n,k;
LL a[1010];
LL b[1010];
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        b[0]=1;
        for(int i=1; i<=k; i++)
            for(int j=i; j<=n; j++)
            {
                a[j]=(a[j-i]+a[j])+(b[j-i]+b[j])/INF;   //高位 进位也在不断更新
                b[j]=(b[j-i]+b[j]) % INF;   //低位

            }
        //当前大小的包里存的种数不断更新

        if(a[n]==0)
            printf("%I64d\n",b[n]);
        else
            printf("%I64d%I64d\n",a[n],b[n]);

    }
    return 0;
}

/*题意：
输入n 和 k
用1到k的数的和凑出 n   问一共有多少种凑发法*/