hdu 3199 丑数2

Hamming Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1124    Accepted Submission(s): 482

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

 

数据 确实挺吓人的     不确定的时候 用long long 或  __int64  

View Code

Problem : 3199 ( Hamming Problem )     Judge Status : Accepted
RunId : 16987110    Language : G++    Author : 1136242673
Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta

#include<stdio.h>#include<string.h>#include<stdlib.h>#define min(a,b)(a<b?a:b)long long a[110000];int p1,p2,p3,n;long long min4(long long a,long long b,long long c){    long long ans1=min(a,b);    long long ans=min(ans1,c);    return ans;}int main(){     int x1,x2,x3;     while(scanf("%d%d%d%d",&p1,&p2,&p3,&n)==4)     {         a[0]=1;         x1=x2=x3=0;         for(int i=1;i<=n;i++)         {             a[i]=min4(a[x1]*p1,a[x2]*p2,a[x3]*p3);             if(a[i]==a[x1]*p1) x1++;             if(a[i]==a[x2]*p2) x2++;             if(a[i]==a[x3]*p3) x3++;         }         printf("%I64d\n",a[n]);     }     return 0;}