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HDU 5015 233 Matrix ( 矩阵快速幂 )
这是西安网络赛的一题，，但是YY之还是没有搞出来。。
后来学习了，今天写个题解吧
题意：
给定一个矩阵的第一列，然后需要推算出第n行第m列的数值
分析： 矩阵快速幂搞之 构造矩阵如下(需要再增加233 和 3 两行进行状态转移) 1 0 0 0 0 0 0 0 0 0 1 0 a 1 1 0 0 0 0 0 0 0 0 1 0 b 1 1 1 0 0 0 0 0 0 0 1 0 c 1 1 1 0 0 0 0 0 0 0 1 0 d 1 1 1 1 0 0 0 0 0 0 1 0 e 1 1 1 1 1 0 0 0 0 0 1 0 f 1 1 1 1 1 1 0 0 0 0 1 0 g 1 1 1 1 1 1 1 0 0 0 1 0 h 1 1 1 1 1 1 1 1 0 0 1 0 i 1 1 1 1 1 1 1 1 1 0 1 0 j 1 1 1 1 1 1 1 1 1 1 1 0 k 0 0 0 0 0 0 0 0 0 0 10 1 233 0 0 0 0 0 0 0 0 0 0 0 1 3
代码君如下
/* *********************************************** Problem :HDU 5015 -- 233 Matrix File Name : Author : Created Time :2014-09-15 13:40:17 ************************************************ */ #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 MOD 10000007 #define MAX_SIZE 12 int m, n; LL M[MAX_SIZE]; inline void scan( LL &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]; int n; Mat(int _n = 0) : n(_n) {}; void zero() { for( int i = 0; i < n; ++i ) for( int j = 0; j < n; ++j ) mat[i][j] = 0; } void init() { for( int i = 0; i < n; ++i ) for( int j = 0; j < n; ++j ) mat[i][j] = ( i == j ); } Mat operator * (const Mat &b) const { Mat c(n); c.zero(); for( int k = 0; k < n; ++k ) for( int i = 0; i < n; ++i ) if( mat[i][k] ) for( int j = 0; j < n; ++j ) c.mat[i][j] = ( c.mat[i][j] + mat[i][k] * b.mat[k][j] ) % MOD; return c; } void debug() { for( int i = 0; i < n; ++i ) { for( int j = 0; j < n; ++j ) { if( j != 0 ) putchar( ‘ ‘ ); printf( "%d", mat[i][j] ); } putchar( ‘\n‘ ); } } }; Mat fast_mod( Mat a, int b ) { Mat res(a.n); res.init(); while( b ) { if( b & 1 ) res = res * a; a = a * a; b >>= 1; } return res; } void Orz() { while( ~scanf( "%d %d", &n, &m ) ) { for( int i = 0; i < n; ++i ) scan( M[i] ); M[n]= 233, M[n+1] = 3; Mat t(n+2); t.zero(); for( int i = 0; i < n; ++i ) for( int j = 0; j <= i; ++j ) t.mat[i][j] = 1; for( int i = 0; i < n; ++i ) t.mat[i][n] = 1; t.mat[n][n] = 10; t.mat[n][n+1] = t.mat[n+1][n+1] = 1; //t.debug(); Mat res(n+2); res = fast_mod( t, m ); //res.debug(); LL ans = 0; for( int i = 0; i < n+2; ++i ) ans = ( ans + res.mat[n-1][i] * M[i] ) % MOD; printf( "%I64d\n", ans ); } } int main() { Orz(); return 0; }
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原文地址：http://www.cnblogs.com/BigBallon/p/4042079.html
