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;
}

 

转载于:https://www.cnblogs.com/sola1994/p/4269041.html