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-12;
const int MAXN = 1005;
const int MAXM = 5005;

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[1610], 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
", time);
//}

void init()
{
    priority_queue<NODE>q;
    NODE U;  U.num=1; U.flag=-1;
    q.push(U);
    int num = 1, tot = 1;
    while(1)
    {
        U = q.top(); q.pop();
        DP[num++] = U.num;
        if(num>1500) return ;
        if(tot > 1600) continue;
        q.push(NODE(U.num*5, 1));                  tot++;
        if(U.flag<=0) {q.push(NODE(U.num*3, 0));   tot++; }
        if(U.flag<=-1){q.push(NODE(U.num*2, -1));  tot++; }
    }
}

int main()
{
   // printf("%d
", (int)(1600 * (log(1600.0)/log(2.0))));
   // freopen("test.in", "r", stdin);
    init();
    while(~scanf("%d", &N) &&N)
    {
        printf("%d
", DP[N]);
    }
    return 0;
}

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[1501];

    i2_mul = 1;
    i3_mul = 1;
    i5_mul = 1;
    ugly[1]=1;

    for(  int i = 2; i <= 1500; i++ )
    {
        ugly[i] = min(ugly[i2_mul]*2,ugly[i3_mul]*3,ugly[i5_mul]*5);
        if(ugly[i] == ugly[i2_mul]*2 )
            i2_mul++;
        if(ugly[i] == ugly[i3_mul]*3 )
            i3_mul++;
        if(ugly[i] == ugly[i5_mul]*5)
            i5_mul++;
    
    }


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

        if( n == 0 )
            break;

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

    }

    return 0;
}