【转】HDU 1058 Humble Numbers:寻找丑数问题?DP?

这个和上一道HDU 3199 Hamming Problem是类似的,有了思路就开始码代码了,可是！序数词的写法不对(代码注释部分)又上网普及了序数词的写法···

搜到其他解题报告 其中有把这道题分类为DP的,最优子结构？无后效性？

还有“寻找丑数问题” 详情点这里。

 

#include<iostream>
#include<cstdio>
using namespace std;
const int Size=5842;
long long num[Size+1];
int main()
{
        int k1, k2, k3, k4, n;
         k1=k2=k3=k4=1;
         num[1]=1;
         for(int i=2; i<Size+1; i++)
         {
                 long long x1=num[k1]*2;
                 long long x2=num[k2]*3;
                 long long x3=num[k3]*5;
                 long long x4=num[k4]*7;

                 long long M1=min(x1, x2);
                 long long M2=min(x3, x4);
                 M1=min(M1, M2);
                 num[i]=M1;
                 if(M1==x1) k1++;
                 if(M1==x2) k2++;
                 if(M1==x3) k3++;
                 if(M1==x4) k4++;
         }
        while(cin>>n&&n!=0)
        {
//                int t=n%10;
//                if(n==11 || n==12 || n==13)
//                    t=n;
                 if(n % 10 == 1 && n % 100 != 11)
                        cout<<"The "<<n<<"st humble number is "<<num[n]<<"."<<endl;
                else if(n % 10 == 2 && n % 100 != 12)
                        cout<<"The "<<n<<"nd humble number is "<<num[n]<<"."<<endl;
                else if(n % 10 == 3 && n % 100 != 13)
                        cout<<"The "<<n<<"rd humble number is "<<num[n]<<"."<<endl;
                else
                        cout<<"The "<<n<<"th humble number is "<<num[n]<<"."<<endl;
//                switch(t)
//                {
//                                case 1: cout<<"The "<<n<<"st humble number is "<<num[n]<<"."<<endl; break;
//                                case 2: cout<<"The "<<n<<"nd humble number is "<<num[n]<<"."<<endl; break;
//                                case 3: cout<<"The "<<n<<"rd humble number is "<<num[n]<<"."<<endl; break;
//                                default:cout<<"The "<<n<<"th humble number is "<<num[n]<<"."<<endl; break;
//                }
        }
        return 0;
}