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：
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <vector> 8 #include <cstdio> 9 #include <cctype> 10 #include <cstring> 11 #include <cstdlib> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 #define INF 0x3f3f3f3f 16 #define MAX(a,b) (a > b ? a : b) 17 #define MIN(a,b) (a < b ? a : b) 18 #define mem0(a) memset(a,0,sizeof(a)) 19 20 typedef long long LL; 21 const double eps = 1e-12; 22 const int MAXN = 1005; 23 const int MAXM = 5005; 24 25 struct NODE 26 { 27 int num; 28 int flag; 29 NODE(){} 30 NODE(int _num, int _flag){num=_num;flag=_flag;} 31 bool operator < (NODE B)const 32 { 33 return num > B.num; 34 } 35 }; 36 int DP[1610], N; 37 38 //void init() 39 //{ 40 // int time = 0; 41 // for(int i=1;i<=1500;i++) DP[i] = INF; 42 // DP[1] = 1; int now = 1; 43 // int two=1, three=1, five=1; 44 // for(int i=2;i<=1500;i++) 45 // { 46 // for(int j=min(two,min(three,five));j<i;j++) 47 // { 48 // time ++; 49 // if(DP[j]*2 > now && DP[i]>DP[j]*2){ DP[i] = min(DP[i], DP[j]*2); two = j;break;} 50 // if(DP[j]*3 > now && DP[i]>DP[j]*3){ DP[i] = min(DP[i], DP[j]*3); three = j;} 51 // if(DP[j]*5 > now && DP[i]>DP[j]*5){ DP[i] = min(DP[i], DP[j]*5); five = j;} 52 // } 53 // now = DP[i]; 54 // } 55 // //printf("Time=%d\n", time); 56 //} 57 58 void init() 59 { 60 priority_queue<NODE>q; 61 NODE U; U.num=1; U.flag=-1; 62 q.push(U); 63 int num = 1, tot = 1; 64 while(1) 65 { 66 U = q.top(); q.pop(); 67 DP[num++] = U.num; 68 if(num>1500) return ; 69 if(tot > 1600) continue; 70 q.push(NODE(U.num*5, 1)); tot++; 71 if(U.flag<=0) {q.push(NODE(U.num*3, 0)); tot++; } 72 if(U.flag<=-1){q.push(NODE(U.num*2, -1)); tot++; } 73 } 74 } 75 76 int main() 77 { 78 // printf("%d\n", (int)(1600 * (log(1600.0)/log(2.0)))); 79 // freopen("test.in", "r", stdin); 80 init(); 81 while(~scanf("%d", &N) &&N) 82 { 83 printf("%d\n", DP[N]); 84 } 85 return 0; 86 }
O(n)的：不是我写的＝ ＝
1 #include <stdio.h> 2 int min(int a,int b,int c) 3 { 4 if(b<a) 5 a=b; 6 if(c<a) 7 a=c; 8 return a; 9 } 10 int main() 11 { 12 int n; 13 int i2_mul; 14 int i3_mul; 15 int i5_mul; 16 unsigned long ugly[1501]; 17 18 i2_mul = 1; 19 i3_mul = 1; 20 i5_mul = 1; 21 ugly[1]=1; 22 23 for( int i = 2; i <= 1500; i++ ) 24 { 25 ugly[i] = min(ugly[i2_mul]*2,ugly[i3_mul]*3,ugly[i5_mul]*5); 26 if(ugly[i] == ugly[i2_mul]*2 ) 27 i2_mul++; 28 if(ugly[i] == ugly[i3_mul]*3 ) 29 i3_mul++; 30 if(ugly[i] == ugly[i5_mul]*5) 31 i5_mul++; 32 33 } 34 35 36 while(true) 37 { 38 scanf("%d",&n); 39 40 if( n == 0 ) 41 break; 42 43 printf("%d\n",ugly[n]); 44 45 } 46 47 return 0; 48 }
