POJ2663,2506 Tri Tiling 递推

POJ2663,2506 Tri Tiling 递推 | 压缩DP

　　题目链接：http://poj.org/problem?id=2663 http://poj.org/problem?id=2506

　　简单的递推题，直接递推过去就可以了。

POJ-2663:

 1 //STATUS:C++_AC_0MS_172KB
 2 #include <functional>
 3 #include <algorithm>
 4 #include <iostream>
 5 //#include <ext/rope>
 6 #include <fstream>
 7 #include <sstream>
 8 #include <iomanip>
 9 #include <numeric>
10 #include <cstring>
11 #include <cassert>
12 #include <cstdio>
13 #include <string>
14 #include <vector>
15 #include <bitset>
16 #include <queue>
17 #include <stack>
18 #include <cmath>
19 #include <ctime>
20 #include <list>
21 #include <set>
22 #include <map>
23 using namespace std;
24 //using namespace __gnu_cxx;
25 //define
26 #define pii pair<int,int>
27 #define mem(a,b) memset(a,b,sizeof(a))
28 #define lson l,mid,rt<<1
29 #define rson mid+1,r,rt<<1|1
30 #define PI acos(-1.0)
31 //typedef
32 typedef __int64 LL;
33 typedef unsigned __int64 ULL;
34 //const
35 const int N=20;
36 const int INF=0x3f3f3f3f;
37 const int MOD=100000,STA=8000010;
38 const LL LNF=1LL<<60;
39 const double EPS=1e-8;
40 const double OO=1e15;
41 const int dx[4]={-1,0,1,0};
42 const int dy[4]={0,1,0,-1};
43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
44 //Daily Use ...
45 inline int sign(double x){return (x>EPS)-(x<-EPS);}
46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
49 template<class T> inline T Min(T a,T b){return a<b?a:b;}
50 template<class T> inline T Max(T a,T b){return a>b?a:b;}
51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
55 //End
56 
57 int f[2][N];
58 int n;
59 
60 int main()
61 {
62  //   freopen("in.txt","r",stdin);
63     int i,j,k,p;
64     while(~scanf("%d",&n) && n!=-1)
65     {
66         mem(f,0);
67         f[0][0]=1;p=1;
68         for(i=0;i<n;i++){
69             for(j=0;j<3;j++,mem(f[p=!p],0)){
70                 for(k=0;k<8;k++){
71                     if(k&(1<<j)){
72                         f[p][k&~(1<<j)]+=f[!p][k];
73                     }
74                     else {
75                         f[p][k|(1<<j)]+=f[!p][k];
76                         if( j<2 && !(k&(1<<(j+1))) ){
77                             f[p][k|(1<<(j+1))]+=f[!p][k];
78                         }
79                     }
80                 }
81             }
82         }
83 
84         printf("%d\n",f[!p][0]);
85     }
86     return 0;
87 }

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POJ-2605（Java大数）：

 1 //STATUS:Java_AC_391MS_5532KB
 2 import java.util.*;
 3 import java.math.*;
 4 import java.io.*;
 5 import java.text.*;
 6 
 7 public class Main {
 8     public static void main(String args[]){
 9         Scanner cin = new Scanner (new BufferedInputStream(System.in));
10         int i,n;
11         BigInteger a,b,c=BigInteger.valueOf(1);
12         while(cin.hasNext()){
13             n=cin.nextInt();
14             a=BigInteger.valueOf(1);
15             b=BigInteger.valueOf(3);
16             if(n==1)c=BigInteger.valueOf(1);
17             if(n==2)c=BigInteger.valueOf(3);
18             for(i=3;i<=n;i++){
19                 c=b.add(a.multiply(BigInteger.valueOf(2)));
20                 a=b;b=c;
21             }
22             System.out.println(c);
23         }
24     }
25 }

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