HDU 3199 Hamming Problem

Hamming Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1305 Accepted Submission(s): 583

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


解析：本题是[HDU 1058 Humble Numbers](http://www.cnblogs.com/inmoonlight/p/6030949.html)的升级版，2、3、5变成了p1、p2、p3。不过原理是相同的，都是运用DP的思想。因为题目有说明所有输入输出小于10^18，我们可以用2、3、5来得到最大可能出现的下标。经过测试，第一个大于或等于10^18的数下标为10916，因而数组的大小最小为10917。


```
#include
#include
#define ll long long
using namespace std;

ll res[11000];

void solve(int a, int b, int c, int n)

{

res[0] = 1;

int p1 = 0, p2 = 0, p3 = 0;

for(int i = 1; i <= n; ++i){

res[i] = min(res[p1]a, min(res[p2]b, res[p3]c));

if(res[i] == res[p1]a) ++p1;

if(res[i] == res[p2]b) ++p2;

if(res[i] == res[p3]c) ++p3;

}

printf("%I64d\n", res[n]);

}

int main()

{

int a, b, c, n;

while(~scanf("%d%d%d%d", &a, &b, &c, &n)){

solve(a, b, c, n);

}

return 0;

}