POJ - 3181 Dollar Dayz

原创于 2019-07-27 21:39:57 发布 · 138 阅读

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本文介绍了一种算法，用于解决农民约翰在商店中用特定金额购买不同价格工具的问题，通过动态规划方法，计算出所有可能的购买组合数量。文章提供了两种实现方案，一种使用二维数组，另一种优化为一维数组，以节省内存。

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

思路：参考如下链接：https://blog.youkuaiyun.com/libin56842/article/details/9455979

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define mod 1000000000000000000
long long a[1010][110];//hou
long long b[1010][110];//qian
int main()
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(int i=0; i<k; i++)
            a[i][0]=1;
        for(int i=1; i<=k; i++)
        {
            for(int j=1; j<=n; j++)
            {
                if(i>j)
                {
                    a[i][j]=a[i-1][j];
                    b[i][j]=b[i-1][j];
                }
                else
                {
                    a[i][j]=(a[i-1][j]+a[i][j-i])%mod;
                    b[i][j]=b[i-1][j]+b[i][j-i]+(a[i-1][j]+a[i][j-i])/mod;
                }
            }
        }
        if(b[k][n])
            printf("%lld%lld\n",b[k][n],a[k][n]);
        else
            printf("%lld\n",a[k][n]);
    }
    return 0;
}

使用一维数组写的，参考如下链接：https://www.cnblogs.com/kuangbin/archive/2012/09/20/2695165.html

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
const int MAXN=1100;
const long long inf=1000000000000000000LL;

long long a[MAXN];//高位
long long b[MAXN];//低位

int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(a));
        b[0]=1;
        for(int i=1;i<=k;i++)
          for(int j=i;j<=n;j++)
          {
              a[j]=a[j]+a[j-i]+(b[j]+b[j-i])/inf;
              b[j]=(b[j]+b[j-i])%inf;
          }
        if(a[n]==0)
        {
            printf("%lld\n",b[n]);
        }
        else
        {
            printf("%lld%lld\n",a[n],b[n]);
        }
    }
    return 0;
}