T解 POJ-3233 [矩阵快速幂][矩阵乘法][二分求解]
2017-03-18 20:12
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大家都很强,可与之共勉
额,第一次写的暴力,果断TLE。然后第二次用二分法,时间过于长。YYF告诉我可以减少初始化次数与mod的次数,虽然我不知道怎么减少mod次数。PS::重定义运算符,比函数慢一些。
Matrix Power Series
Time Limit: 3000MS Memory Limit: 131072K
Total Submissions: 11954 Accepted: 5105
Description
Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.
Input
The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 109) and m (m < 104). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.
Output
Output the elements of S modulo m in the same way as A is given.
Sample Input
2 2 4
0 1
1 1
Sample Output
1 2
2 3
#include "cctype" #include "cstdio" #include "cstring" template <class T> inline bool readIn(T &x) { x = 0; T flag = 1; char ch = (char)getchar(); while (!isdigit(ch)) { if (ch == '-') flag = -1; ch = (char)getchar(); } while (isdigit(ch)) x = (x << 1) + (x << 3) + ch - 48, ch = (char)getchar(); x *= flag; } template <class T> inline void write(T x) { if (x > 9) write(x / 10); putchar(x % 10 + 48); } template <class T> inline bool writeIn(T x) { if( x < 0 ) { putchar('-'); x = -x; } write(x); putchar(' '); } int n, k, m; struct Matrix { int m[30][30]; inline bool unit() { memset(m, 0, sizeof(m)); for (int i = 0; i < n; m[i][i] = 1, ++i); } inline bool empty() { memset(m, 0, sizeof(m)); } } a; Matrix operator * ( Matrix a, Matrix b ) { Matrix c; c.empty(); for(int i = 0; i < n; ++i) for(int j = 0; j < n; ++j) for(int k = 0; k < n; ++k) c.m[i][j] = (c.m[i][j] + a.m[i][k] * b.m[k][j]) % m; return c; } Matrix operator + ( Matrix a, Matrix b ) { Matrix c; for(int i = 0; i < n; ++i) for(int j = 0; j < n; ++j) c.m[i][j] = (a.m[i][j] + b.m[i][j]) % m; return c; } inline Matrix pow( Matrix a, int x ) { Matrix rt; for (rt.unit(); x; (x & 1) ? rt = rt * a : rt, x >>= 1, a = a * a); return rt; } Matrix sum ( Matrix s , int k ) { if ( k == 1 ) return s ; Matrix tmp ; tmp.unit(); tmp = tmp + pow( s , k >> 1 ); tmp = tmp * sum( s , k >> 1 ); if ( k & 1 ) tmp = tmp + pow( s , k ); return tmp ; } int main() { readIn(n), readIn(k), readIn(m); for(int i = 0; i < n; ++i) for(int j = 0; j < n; ++j) readIn(a.m[i][j]); a = sum(a, k); for(int i = 0; i < n; printf("\n"), ++i) for(int j = 0; j < n; ++j) writeIn(a.m[i][j]); }
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