POJ-3420 Quad Tiling 状态压缩+矩阵乘法

　　题目链接：http://poj.org/problem?id=3420

　　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 }