hdu 1058 Humble Numbers || poj 1338（dp）

 

两题都是一样的题目 只是hdu 1058 多了个7

 

题意：求一个每个数因子仅含2 3 5 7 的 序列

         问 第n个数是几

 

思路：

        ans[i]=min(min(ans[n2]*2,ans[n3]*3),min(ans[n5]*5,ans[n7]*7)); if(ans[i]==ans[n2]*2) n2++; if(ans[i]==ans[n3]*3) n3++; if(ans[i]==ans[n5]*5) n5++; if(ans[i]==ans[n7]*7) n7++;
 

hdu 1058

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

int INF=2000000005;

int ans[6000];

int main()

{

    int coun=2;

    ans[1]=1;

    int n,i,j,k;

    int n2=1,n3=1,n5=1,n7=1;

    for(i=2;i<=5842;i++)

    {

        ans[i]=min(min(ans[n2]*2,ans[n3]*3),min(ans[n5]*5,ans[n7]*7));

        if(ans[i]==ans[n2]*2) n2++;

        if(ans[i]==ans[n3]*3) n3++;

        if(ans[i]==ans[n5]*5) n5++;

        if(ans[i]==ans[n7]*7) n7++;

    }

    while(scanf("%d",&n),n)

    {

        if(n%10==1&&n%100!=11)

        printf("The %dst humble number is %d.\n",n,ans[n]);

        else if(n%10==2&&n%100!=12)

        printf("The %dnd humble number is %d.\n",n,ans[n]);

        else if(n%10==3&&n%100!=13)

        printf("The %drd humble number is %d.\n",n,ans[n]);

        else

        printf("The %dth humble number is %d.\n",n,ans[n]);

    }

    return 0;

}