HDU 1058 Humble Numbers

http://acm.hdu.edu.cn/showproblem.php?pid=1058

话说这题是动态规划的吧. 刚开始就想到的是直接暴力, 用一个有序数列.

数列的最开头的数字是最小的,然后将它乘以2,3,5,7后得到的数字放回数列中,同时将第一个数字出列....

然后只要出了5842个数字就刚好了.

set 容器刚好满足 1.去重, 2.排序  

#include <stdio.h>
#include <iostream>
#include <set>
using namespace std;
const int maxn = 5843;
set<long long> ans;
long long temp[maxn];
int main()
{
    int i, j, n;
    long long k;
    temp[1]=1;
    ans.insert(2);
    ans.insert(3);
    ans.insert(5);
    ans.insert(7);
    set<long long>::iterator iter;
    for(i=2;i<=5842;i++)
    {
        iter = ans.begin();
        temp[i] = *iter;
        k = *iter;
        ans.insert(2*k);
        ans.insert(3*k);
        ans.insert(5*k);
        ans.insert(7*k);
        ans.erase(iter);
    }
    while(~scanf("%d",&n),n){
        if(n%100==11 || n%100==12 || n%100==13){
            printf("The %dth humble number is %lld.\n",n,temp[n]);
        }
        else if(n%10==1){
            printf("The %dst humble number is %lld.\n",n,temp[n]);
        }
        else if(n%10==2){
            printf("The %dnd humble number is %lld.\n",n,temp[n]);
        }
        else if(n%10==3){
            printf("The %drd humble number is %lld.\n",n,temp[n]);
        }
        else{
            printf("The %dth humble number is %lld.\n",n,temp[n]);
        }
    }
    return 0;
}