Matrix Power Series(矩阵快速幂)

矩阵快速幂:http://www.cnblogs.com/kuangbin/archive/2012/08/17/2643347.html

Matrix Power Series

Time Limit: 3000MS Memory Limit: 131072KTotal Submissions: 16341 Accepted: 6966

Description

Given a n × n matrix A and a positive integer k, find the sumS =A + A² + A³ + … + A^k.

Input

The input contains exactly one test case. The first line of input contains three positive integersn (n ≤ 30),k (k ≤ 10⁹) and m (m < 10⁴). Then follown lines each containingn nonnegative integers below 32,768, givingA’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 40 11 1

Sample Output

1 22 3

题意:简单易懂;

题解:两次二分矩阵,具体看代码吧!

AC代码:

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <vector>#include <cmath>#include <queue>#include <map>#include <set>#define eps 1e-9using namespace std;typedef long long ll;typedef pair<int,int>P;const int M = 1e5 + 100;const int INF = 0x3f3f3f3f;int n,mod,k;struct Mac {    int c[33][33]; //原始矩阵    void unit1() { //原始矩阵        for(int i = 0; i < n; i++) {            for(int j = 0; j < n; j++) {                scanf("%d",&c[i][j]);            }        }    }    void unit2() { //单位矩阵        c[0][0] = c[1][1] = 1;        c[1][0] = c[0][1] = 0;    }};Mac tmp,sum; //原始矩阵,和求和矩阵Mac mul(Mac a,Mac b) { //矩阵相乘    Mac p;    memset(p.c,0,sizeof(p.c));    for(int i = 0; i < n; i++) {        for(int j = 0; j < n; j++) {            if(a.c[i][j]) {                for(int k = 0; k < n; k++) {                    p.c[i][k] += (a.c[i][j] * b.c[j][k]) % mod;                    p.c[i][k] %= mod;                }            }        }    }    return p;}Mac add(Mac a,Mac b) { //矩阵加法    for(int i = 0; i < n; i++) {        for(int j = 0; j < n; j++) {            a.c[i][j] += b.c[i][j] % mod;            a.c[i][j] %= mod;        }    }    return a;}Mac quickmod(Mac a,int n) { //矩阵快速幂    Mac p;    p.unit2();    while(n) {        if(n & 1) p = mul(p,a);        a = mul(a,a);        n >>= 1;    }    return p;}Mac binary_matrix1(int k) { //二分矩阵快速幂    Mac res;    if(k == 1) {        return tmp;    }    res = binary_matrix1(k / 2);    res = mul(res,res);    if(k & 1) res = mul(res,tmp);    return res;}Mac binary_matrix2(int k){ //二分矩阵求矩阵幂之和 A1+A2+A3...;    if(k == 1){        return tmp;    }    Mac res = binary_matrix2(k / 2);    Mac tp = binary_matrix1(k / 2);    tp = mul(tp,res);    sum = add(sum,tp);    if(k & 1) sum = add(sum,binary_matrix1(k));    return sum;}int main() {    cin>>n>>k>>mod;    tmp.unit1();    for(int i = 0; i < n; i++){        for(int j = 0; j < n; j++){            sum.c[i][j] = tmp.c[i][j];        }    }    Mac res = binary_matrix2(k);    for(int i = 0; i < n; i++) {        for(int j = 0; j < n; j++) {            printf(j == n - 1 ? "%d\n" : "%d ",res.c[i][j]);        }    }    return 0;}