FZU 1911 Construct a Matrix ( 矩阵快速幂 + 构造 )

FZU 1911 Construct a Matrix (  矩阵快速幂 + 构造 )

题意：

需要构造一个矩阵满足如下要求：

1.矩阵是一个S(N)*S(N)的方阵
2.S(N)代表斐波那契数列的前N项和模上M
3.矩阵只能由1, 0, -1组成
4.矩阵每行每列的和不能相等
Here, the Fibonacci numbers are the numbers in the following
sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …

分析：

1.可以发现，如果S(N) 为奇数，无法构造，直接输出No
2.可以用一个3*3的矩阵通过矩阵快速幂来解决斐波那契数列前N项和问题
3.4. 这个倒是真的不知道怎么构造。。。

FZU大神的构造方法，，太牛了

上三角全为1
下三角全为-1
对角线为 1 0 交替

S(N) = 4，构造如下

1  1  1  11  1  0 -11  1 -1 -10 -1 -1 -1

代码

/* ***********************************************
Problem       :FZU 1911 Construct a Matrix
File Name     :
Author        :
Created Time  :2014-10-21 13:58:53
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <map>
#include <string>
#include <set>
#include <algorithm>
#include <cmath>
#include <ctime>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define CLR(a,b)     memset( a, b, sizeof(a) )
#define MAX_SIZE 3

int m, n, r;
int M[MAX_SIZE];
inline void scan( int &x )
{
    char c;
    while( c = getchar(), c < ‘0‘ || c > ‘9‘ );
    x = c - ‘0‘;
    while( c = getchar(), c >= ‘0‘ && c <= ‘9‘ ) x = x*10 + c - ‘0‘;
}

struct Mat
{
    LL mat[MAX_SIZE][MAX_SIZE];
    void zero()
    {
        for( int i = 0; i < 3; ++i )
            for( int j = 0; j < 3; ++j )
                mat[i][j] = 0;
    }
    void init()
    {
        for( int i = 0; i < 3; ++i )
            for( int j = 0; j < 3; ++j )
                mat[i][j] = ( i == j );
    }
    void setv( int v )
    {
        for( int i = 0; i < 3; ++i )
            for( int j = 0; j < 3; ++j )
                mat[i][j] = v;
    }
    Mat operator * (const Mat &b) const
    {
        Mat c;
        c.zero();
        for( int k = 0; k < 3; ++k )
            for( int i = 0; i < 3; ++i ) if( mat[i][k] )
                for( int j = 0; j < 3; ++j )
                    c.mat[i][j] = ( c.mat[i][j] + mat[i][k] * b.mat[k][j] ) % m;
        return c;
    }
    void debug()
    {
        for( int i = 0; i < 3; ++i )
        {
            for( int j = 0; j < 3; ++j )
            {
                if( j != 0 )    putchar( ‘ ‘ );
                printf( "%lld", mat[i][j] );
            }
            putchar( ‘\n‘ );
        }
    }
};

Mat fast_mod( Mat a, int b )
{
    Mat res;
    res.init();
    while( b )
    {
        if( b & 1 )   res = res * a;
        a = a * a;
        b >>= 1;
    }
    return res;
}

void Orz()
{
    //freopen( "1.txt", "r", stdin );
    //freopen( "2.txt", "w", stdout );
    int cas = 0, t;
    scan( t );
    while( t-- )
    {
        scan( n ); scan( m );
        if( n == 1 )
            r = 1;
        else if( n == 2 )
            r = 2 % m;
        else
        {
            Mat c;
            c.setv(1);
            c.mat[0][0] = c.mat[0][2] = c.mat[1][2] = 0;
            //c.debug();
            c = fast_mod( c, n-2 );
            //c.debug();
            r = 0;
            M[0] = M[1] = 1, M[2] = 2;

            for( int i = 0; i < 3; ++i )
                r = ( r + c.mat[2][i] * M[i] ) % m ;
        }
        printf( "Case %d: ", ++cas );
        if( r == 0 || r & 1 )
            puts( "No" );
        else
        {
            puts( "Yes" );
            int out[210][210];
            for( int i = 0; i < r; ++i )
            {
                for( int j = 0; j < r-i; ++j )
                {
                    out[i][j] = 1;
                }
            }
            for( int i = 0; i < r; ++i )
            {
                for( int j = 0; j < i; ++j )
                {
                    out[i][r-1-j] = -1;
                }
            }
            int flag = 1;
            for( int i = 0; i < r; ++i )
            {
                for( int j = 0; j < r; ++j )
                {
                    if( i == r - j - 1 )
                    {
                        if( flag )    out[i][j] = 1;
                        else        out[i][j] = 0;
                        flag ^= 1;
                    }
                }
            }
            for( int i = 0; i < r; ++i )
            {
                for( int j = 0; j < r ; ++j )
                {
                    if( j != 0 ) putchar( ‘ ‘ );
                    printf( "%d", out[i][j] );
                }
                putchar( ‘\n‘ );
            }

        }
    }
}

int main()
{
    Orz();
    return 0;
}

代码君