Hamming Problem（丑数）

蔚落 2022-08-05 02:46 119阅读 0赞

# Hamming Problem #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 1002    Accepted Submission(s): 422**

Problem Description

For each three prime numbers p1, p2 and p3, let's define Hamming sequence Hi(p1, p2, p3), i=1, ... as containing in increasing order all the natural numbers whose only prime divisors are p1, p2 or p3.  
  
For example, H(2, 3, 5) = 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, ...  
  
So H5(2, 3, 5)=6.

Input

In the single line of input file there are space-separated integers p1 p2 p3 i.

Output

The output file must contain the single integer - Hi(p1, p2, p3). All numbers in input and output are less than 10^18.

Sample Input

7 13 19 100

Sample Output

26590291

Source

[Northeastern Europe 2000 - Far-Eastern Subregion][]

这个得有丑数的概念才能写！

什么是丑数？

描述：把只包含因子2、3和5的数称作丑数（Ugly Number）。例如6、8都是丑数，但14不是，因为它包含因子7。习惯上我们把1当做是第一个丑数。求按从小到大的顺序的第N个丑数。

#include<cstdio>
    #include<iostream>
    #include<cstring>
    using namespace std;
    __int64 dp[100001];
    __int64 min3(__int64 a,__int64 b,__int64 c)
    {
        a=min(a,b);
        a=min(a,c);
        return a;
    }
    int main()
    {
        __int64 a,b,c,d;
        while(scanf("%I64d%I64d%I64d%I64d",&a,&b,&c,&d)!=EOF)
        {
            dp[0]=1;
            __int64 aa=0,bb=0,cc=0;
            __int64 aaa,bbb,ccc;
            for(__int64 i=1;i<=d;i++)
            {
                dp[i]=min3(aaa=dp[aa]*a,bbb=dp[bb]*b,ccc=dp[cc]*c);
                if(dp[i]==aaa)
                aa++;
                if(dp[i]==bbb)
                bb++;
                if(dp[i]==ccc)
                cc++;
            }
            printf("%I64d\n",dp[d]);
        }
    }

[Northeastern Europe 2000 - Far-Eastern Subregion]: http://acm.hdu.edu.cn/search.php?field=problem&key=Northeastern+Europe+2000++-++Far-Eastern+Subregion&source=1&searchmode=source