还是一样的题，，，不解释。不过数据真的太弱了，竟然可以0ms过，，，，看来今天真的很水。。。。。。题目：
Hamming Problem
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5665 | Accepted: 2546 |
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.
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
26590291ac代码：
#include <iostream>
#include <cstdio>
using namespace std;
long long min(long long a,long long b,long long c){long long x=a<b?a:b;return x<c?x:c;
}
int main(){long long prime1,prime2,prime3,order;while(~scanf("%lld%lld%lld%lld",&prime1,&prime2,&prime3,&order)){long long num[100005];num[1]=1;int i=1,p1,p2,p3;p1=p2=p3=1;while(i<=order+1){num[++i]=min(prime1*num[p1],prime2*num[p2],prime3*num[p3]);if(num[i]==prime1*num[p1]) p1++;if(num[i]==prime2*num[p2]) p2++;if(num[i]==prime3*num[p3]) p3++;}printf("%lld\n",num[order+1]);}return 0;
}
