POJ-2663,2506 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 }

View Code

 

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 }

View Code