How many Knight Placing? UVA - 11091

How many Knight Placing?

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description

 

I I U P C 2 0 0 6
Problem H: How many Knight Placing?
Input: standard input Output: standard output
You are given a 6*n chessboard. Yes it is not a regular chessboard. The number of columns in this chessboard is variable. In each of the columns you have to place exactly 2 knights. So you have to place total 2*n knights. You have to count the number of valid placing of these 2*n knight. A placing is invalid if any of the 2 knights attack each other. Those who are not familiar with knight moves “A knight in cell(x,y) attacks the knights in the cell(x±2,y±1) and cell(x±1,y±2)”.
Input
The first line of the input contains a single integer T indicating the number of test cases.  Each test case contains a single integer n.
Output
For each test case output an integer the number of valid placing. The integer may be very large. So just output the result%10007.
Constraints
-           T ≤ 15 -           1 ≤ n < 1000000000
Sample Input	Output for Sample Input
4 1 10 100 1000	15 178 30 8141

好恶心的矩阵构造啊！

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 using namespace std;
 #define mod 10007
 typedef struct point
 {
     int x,y;
 } point;
 typedef struct mar
 {
     int a[][];
 } mar;
 mar ri,anss,riri;
 point p[];
 int a[];
 int check(int t)
 {
     int sum=;
     while(t)
     {
         if(t&)sum++;
         t>>=;
     }
     return sum==;
 }
 bool fun(int x,int y)
 {
     if((x<<)&y||(y<<)&x)return ;
     return ;
 }
 bool fun1(int x,int y,int z)
 {
     if(!fun(x,y)||!fun(y,z))return ;
     if((x<<)&z||(z<<)&x)return ;
     return ;
 }
 mar mul(mar x,mar y)
 {
     mar z;
     memset(z.a,,sizeof(z.a));
     int i,j,k;
     for(k=; k<; k++)
         for(i=; i<; i++)
             if(x.a[i][k])
                 for(j=; j<; j++)
                     z.a[i][j]+=x.a[i][k]*y.a[k][j]%mod,z.a[i][j]%=mod;
     return z;
 }
 void init()
 {
     int tt=(<<)-,nu=,i,j;
     while(tt)
     {
         if(check(tt))
             a[nu++]=tt;
         tt--;
     }
     nu=;
     for(i=; i<; i++)
         for(j=; j<; j++)
             if(fun(a[i],a[j]))
                 p[nu].x=a[i],p[nu++].y=a[j];
     memset(ri.a,,sizeof(ri.a));
     for(i=; i<; i++)
     {
         for(j=; j<; j++)
         {
             if(p[i].y==p[j].x)
             {
                 if(fun1(p[i].x,p[i].y,p[j].y))
                     ri.a[i][j]=;
             }
         }
     }
 }
 int solve(int m)
 {
     int i,j,ans=;
     memset(anss.a,,sizeof(anss.a));
      memset(riri.a,,sizeof(riri.a));
     for(i=; i<; i++)anss.a[i][i]=;
     for(i=; i<; i++)
         for(j=; j<; j++)riri.a[i][j]=ri.a[i][j];
     m-=;
     while(m)
     {
         if(m&)
             anss=mul(anss,riri);
         riri=mul(riri,riri);
         m>>=;
     }
     for(i=; i<; i++)
         for(j=; j<; j++)ans+=anss.a[i][j],ans%=mod;
     return ans;
 }
 int main()
 {
     init();
     int t,m;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&m);
         if(m==)
         {
             printf("15\n");
             continue;
         }
         else if(m==)
         {
             printf("69\n");
             continue;
         }
         printf("%d\n",solve(m));
     }
 }