hdu 3117 矩阵快速幂

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求前四位的时候直接用公式来求

求和四位的时候利用矩阵快速幂来求，每次对10000取余即可

AC代码如下：

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>using namespace std;const int MAX_N = 3;const int MOD = 10000;int fib[40];int N, M;int get_f4( int n ){    double t1 = ( 1 + sqrt( 5.0 ) ) / 2;    double t2 = -0.5 * log10(5.0) + (double)n * log10( t1 );    double t3 = t2 - (int)t2;    int ans = pow( 10.0, t3 ) * 1000.0;    return ans;}void multipy( int a[MAX_N][MAX_N], int b[MAX_N][MAX_N], int c[MAX_N][MAX_N] ){    for( int i = 1; i <= 2; i++ ){        for( int j = 1; j <= 2; j++ ){            c[i][j] = 0;            for( int k = 1; k <= 2; k++ ){                c[i][j] = ( c[i][j] + a[i][k] * b[k][j] ) % MOD;            }        }    }}void get_matrix_pow( int a[MAX_N][MAX_N], int n ){    int ans[MAX_N][MAX_N] = {0};    int temp[MAX_N][MAX_N];    for( int i = 1; i <= 2; i++ )   ans[i][i] = 1;    while( n ){        if( n % 2 == 1 ){            multipy( ans, a, temp );            memcpy( ans, temp, sizeof( int ) * MAX_N * MAX_N );        }        multipy( a, a, temp );        memcpy( a, temp, sizeof( int ) * MAX_N * MAX_N );        n /= 2;    }    memcpy( a, ans, sizeof( int ) * MAX_N * MAX_N );}int main(){    fib[0] = 0;fib[1] = 1;    for( int i = 2; i < 40; i++ ){        fib[i] = fib[i-1] + fib[i-2];    }    while( scanf( "%d", &M ) != EOF ){        if( M < 40 ){            printf( "%d\n", fib[M] );        }else{            int ans_a = get_f4( M );            int a[MAX_N][MAX_N];            a[1][1] = 1;a[1][2] = 1;            a[2][1] = 1;a[2][2] = 0;            get_matrix_pow( a, M );            int ans_b = a[2][1];            printf( "%d...%04d\n", ans_a, ans_b );        }    }    return 0;}

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