POJ3420 Quad Tiling 状态压缩+矩阵乘法

　　非常经典的题目，推荐看<十个利用矩阵乘法解决的经典题目>。先求出相邻两列的状态转移矩阵，然后用矩阵乘法优化，相当于求在一个图上求两点之间有多少条路径数。
  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 const int size=6;
 58 LL ma[size][size]={
 59 {1,1,0,1,1,1},
 60 {1,0,0,0,1,0},
 61 {0,0,0,1,0,0},
 62 {1,0,1,0,0,0},
 63 {1,1,0,0,0,0},
 64 {1,0,0,0,0,0}
 65 };
 66 LL n,MOD;
 67 
 68 struct Matrix{
 69     LL ma[size][size];
 70     Matrix friend operator * (const Matrix a,const Matrix b){
 71         Matrix ret;
 72         mem(ret.ma,0);
 73         int i,j,k;
 74         for(k=0;k<size;k++)
 75             for(i=0;i<size;i++)
 76                 for(j=0;j<size;j++)
 77                     ret.ma[i][j]=(ret.ma[i][j]+a.ma[i][k]*b.ma[k][j])%MOD;
 78         return ret;
 79     }
 80 }ans,mta;
 81 
 82 void mutilpow(LL k)
 83 {
 84     int i,j;
 85     mem(ans.ma,0);
 86     for(i=0;i<size;i++)
 87         ans.ma[i][i]=1;
 88     for(;k;k>>=1){
 89         if(k&1)ans=ans*mta;
 90         mta=mta*mta;
 91     }
 92 }
 93 
 94 int main()
 95 {
 96  //   freopen("in.txt","r",stdin);
 97     int i,j;
 98     while(~scanf("%I64d%I64d",&n,&MOD) && (n||MOD))
 99     {
100         for(i=0;i<size;i++){
101             for(j=0;j<size;j++)
102                 mta.ma[i][j]=ma[i][j];
103         }
104 
105         mutilpow(n);
106 
107         printf("%I64d\n",ans.ma[0][0]);
108     }
109     return 0;
110 }