P2723 丑数 Humble Numbers

　　I really thought out ...... oh ...... Do not blame me

　　I think that has been the number of plug to the pile, know the number of the number of super multi n lot.

　　First of all, a lot more than this is difficult to control, secondly, the degree of complexity is not allowed.

　　So we consider a wave of dp:

　　f [i] denotes the i-th ugly number, then the number of ugly f [i] must be equal to f [k] * a [j] (where k <i).

　　So we have for each i, we enumerate k and j, is greater than f [i - 1] to find the smallest number that is f [i].

　　But this is the n- ² * k, for the 1e5 ...... we have to consider optimization.

　　Then consider a wave:

　　Suppose f [i] = f [k] * a [j] 

　　Suppose f [i + 1] = f [t] * a [j] 

　　Then t must be greater than k, so for each j, k we have now is how much maintenance, start looking for each time from k + 1.

　　Such complexity becomes n * k + k * n (well understood), i.e. n * k.

　　And, long long ......

　　Code:

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define maxn 400005
#define int long long
#define inf 99999999999999
int p[maxn],f[maxn],s[maxn];
main()
{
    int k,n;
    scanf("%lld%lld",&k,&n);
    for(int i=1;i<=k;i++)
        scanf("%lld",&p[i]);
    f[0]=1;
    for(int i=1;i<=n;i++)
    {
        int MIN=inf;
        for(int j=1;j<=k;j++)
        {
            while(f[i-1]>=f[s[j]]*p[j]) s[j]++;
            MIN=min(MIN,f[s[j]]*p[j]);
        }
        f[i]=MIN;
    }
    printf("%lld\n",f[n]);
    return 0;
}