POJ1338Ugly Numbers（DP）

http://poj.org/problem?id=1338

第一反应就是DP，DP[i] = min{2*DP[j], 3*DP[k], 5*DP[p] j,k,p<i};于是枚举一下0～i－1即可

后来听到室友说，可以通过上一个×2、×3、×5得到。于是搞了个优先队列预处理。

后来看了一下以前A的代码。O(n)的。。。（虽然不是我自己做出的＝ ＝）

时间都是0Ms

O(n^2)和O(nlogn)的：

 #include <map>
 #include <set>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <ctime>
 #include <vector>
 #include <cstdio>
 #include <cctype>
 #include <cstring>
 #include <cstdlib>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 #define INF 0x3f3f3f3f
 #define MAX(a,b) (a > b ? a : b)
 #define MIN(a,b) (a < b ? a : b)
 #define mem0(a) memset(a,0,sizeof(a))

 typedef long long LL;
 const double eps = 1e-;
 const int MAXN = ;
 const int MAXM = ;

 struct NODE
 {
     int num;
     int flag;
     NODE(){}
     NODE(int _num, int _flag){num=_num;flag=_flag;}
     bool operator < (NODE B)const
     {
         return num > B.num;
     }
 };
 int DP[], N;

 //void init()
 //{
 //    int time = 0;
 //    for(int i=1;i<=1500;i++) DP[i] = INF;
 //    DP[1] = 1;  int now = 1;
 //    int two=1, three=1, five=1;
 //    for(int i=2;i<=1500;i++)
 //    {
 //        for(int j=min(two,min(three,five));j<i;j++)
 //        {
 //            time ++;
 //            if(DP[j]*2 > now && DP[i]>DP[j]*2){ DP[i] = min(DP[i], DP[j]*2); two = j;break;}
 //            if(DP[j]*3 > now && DP[i]>DP[j]*3){ DP[i] = min(DP[i], DP[j]*3); three = j;}
 //            if(DP[j]*5 > now && DP[i]>DP[j]*5){ DP[i] = min(DP[i], DP[j]*5); five = j;}
 //        }
 //        now = DP[i];
 //    }
 //    //printf("Time=%d\n", time);
 //}

 void init()
 {
     priority_queue<NODE>q;
     NODE U;  U.num=; U.flag=-;
     q.push(U);
     int num = , tot = ;
     while()
     {
         U = q.top(); q.pop();
         DP[num++] = U.num;
         if(num>) return ;
         if(tot > ) continue;
         q.push(NODE(U.num*, ));                  tot++;
         if(U.flag<=) {q.push(NODE(U.num*, ));   tot++; }
         if(U.flag<=-){q.push(NODE(U.num*, -));  tot++; }
     }
 }

 int main()
 {
    // printf("%d\n", (int)(1600 * (log(1600.0)/log(2.0))));
    // freopen("test.in", "r", stdin);
     init();
     while(~scanf("%d", &N) &&N)
     {
         printf("%d\n", DP[N]);
     }
     return ;
 }

O(n)的：不是我写的＝ ＝

 #include <stdio.h>
 int min(int a,int b,int c)
 {
     if(b<a)
     a=b;
     if(c<a)
     a=c;
     return a;
 }
 int main()
 {
     int n;
     int i2_mul;
     int i3_mul;
     int i5_mul;
     unsigned long ugly[];

     i2_mul = ;
     i3_mul = ;
     i5_mul = ;
     ugly[]=;

     for(  int i = ; i <= ; i++ )
     {
         ugly[i] = min(ugly[i2_mul]*,ugly[i3_mul]*,ugly[i5_mul]*);
         if(ugly[i] == ugly[i2_mul]* )
             i2_mul++;
         if(ugly[i] == ugly[i3_mul]* )
             i3_mul++;
         if(ugly[i] == ugly[i5_mul]*)
             i5_mul++;

     }

     while(true)
     {
         scanf("%d",&n);

         if( n ==  )
             break;

         printf("%d\n",ugly[n]);

     }

     return ;
 }