poj 3233 矩阵乘方

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// A^n.cpp : 定义控制台应用程序的入口点。////2分法求矩阵冥次//Poj 3233#include "stdafx.h"#include <iostream>#include <algorithm>using namespace std;//如果矩阵过大或者冥过多都会造成栈溢出const int MAXN = 31;int Mod;typedef struct Matrix{int s[MAXN][MAXN];}M;int n;//a+bM Add(M a, M b){M c;for(int i=1; i<=n; i++){for(int j=1; j<=n; j++){c.s[i][j] = (a.s[i][j] + b.s[i][j]) % Mod;}}return c;}//a*bM Multiply(M a, M b){M c;memset(c.s,0,sizeof(c.s));for(int i=1; i<=n; i++){for(int j=1; j<=n; j++){for(int k=1; k<=n; k++){c.s[i][j] = (c.s[i][j] + (a.s[i][k] * b.s[k][j] % Mod))%Mod ;}}}return c;}//a^tM Pow(M a, int t){M b;if(t == 0){memset(b.s,0,sizeof(b.s));for(int i=1; i<=n; i++){b.s[i][i] = 1;}return b;}else{M k = Pow(a,t/2);if(t&1) //t为奇数{return Multiply(Multiply(k,k),a);}else{return Multiply(k,k);}}}//A + A^2 + A^3 + ... + A^tM Sum(M a, int t){M b;if(t == 1){return a;}else{M k = Sum(a,t/2);if(t&1){M p = Pow(a,t/2+1);return Add (Add(k,p) , Multiply(k,p) );}else{return Add(k , Multiply(Pow(a,t/2),k));}}}int main(){M a;int t;cin>>n>>t>>Mod;for(int i=1; i<=n; i++){for(int j=1; j<=n; j++){cin>>a.s[i][j];a.s[i][j] %= Mod;}}a = Sum(a,t);for(int i=1; i<=n; i++){for(int j=1; j<=n; j++){cout<<a.s[i][j];if(j != n)cout<<" ";}cout<<endl;}return 0;}