【poj3233】Matrix Power Series 矩阵+快速幂

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 40 11 1

Sample Output

1 22 3

Source

POJ Monthly–2007.06.03, Huang, Jinsong

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题意：给你一个n*n的矩阵A，求

∑k=1nAｋ

其中每个元素mod m。

二分，可以像快速幂一样递归处理。

先把A想象成数字，则：

A1+A2+A3+A4+A5+A6

=(A1+A2+A3)+A3∗(A1+A2+A3)

=(1+A3)∗(A1+A2+A3)

对于矩阵，重载了+和*，1为单位矩阵，就一样做了。

代码（不知道为何，很慢）：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int n,mod;struct matrix{    int num[32][32];    void init()     {         memset(num,0,sizeof(num));        for(int i = 1;i <= n;i ++) num[i][i] = 1;    }};matrix operator *(const matrix &a,const matrix &b){    matrix ans;    for(int i = 1;i <= n;i ++)    {        for(int j = 1;j <= n;j ++)        {            ans.num[i][j] = 0;             for(int k = 1;k <= n;k ++)            {                ans.num[i][j] += a.num[i][k] * b.num[k][j];                ans.num[i][j] %= mod;            }        }    }    return ans;}matrix operator +(const matrix &a,const matrix &b){    matrix ans;    for(int i = 1;i <= n;i ++)    {        for(int j = 1;j <= n;j ++)        {            ans.num[i][j] = (a.num[i][j] + b.num[i][j]) % mod;        }    }    return ans;}matrix ksm(matrix a,int b){    matrix ans;    ans.init();    while(b)    {        if(b & 1) ans = ans * a;        a = a * a;        b >>= 1;    }    return ans;}void print(const matrix &ans){    for(int i = 1;i <= n;i ++)    {        for(int j = 1;j <= n;j ++)        {            printf("%d ",ans.num[i][j]);        }        puts("");    }       }matrix ask(const matrix &a,int b){    if(b == 1) return a;    matrix ans;    ans.init();    ans = (ans + ksm(a,b >> 1)) * ask(a,b >> 1);//  print(ans);    if(b & 1)         return ans + ksm(a,b);    else         return ans;}int main(){    matrix ans;    int k;    while(~scanf("%d%d%d",&n,&k,&mod))    {        for(int i = 1;i <= n;i ++)        {            for(int j = 1;j <= n;j ++)            {                scanf("%d",&ans.num[i][j]);            }        }        ans = ask(ans,k);        print(ans);    }    return 0;}