Dollar Dayz POJ - 3181 动态规划完全背包

向右看齐 2022-06-17 11:41 120阅读 0赞

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*/