POJ 3233 Matrix Power Series [矩阵快速幂]【数论】[水]

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题目链接 : http://poj.org/problem?id=3233

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Matrix Power Series
Time Limit: 3000MS Memory Limit: 131072K
Total Submissions: 20930 Accepted: 8760
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
Source
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题目大意:
不用解释了吧就是求Sn
Sn = A + A^2 + A^3 + … + A^k.

解题思路：
这种题目一定想到矩阵快速幂
然后就是怎么构造矩阵了

[A O] 乘 [A E] 等 [A^2 S1]
[O O] 号 [O E]号 [O S0]

矩阵大致就是这么构造出来的

然后注意的事E只有主对角线是1 剩下的都是0 (Sb的我全写成E然后WA的都怀疑人生了)

附本题代码
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#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;using namespace std;typedef long long int LL ;#define INF 0x3f3f3f3f#define pb push_back#define lalal puts("*******");/*************************************/int MOD;const int M = 32*2;struct Matrix{    LL m[M][M];    void display(int N)    {        for(int i=0; i<N; i++)        {            for(int j=0; j<N; j++)            {                if(j) printf(" ");                printf("%I64d",m[i][j]);            }            puts("");        }    }    void clearI()    {        for(int i=0; i<M; i++)            for(int j=0; j<M; j++)                m[i][j]=(i==j);    }    void clearO()    {        for(int i=0; i<M; i++)            for(int j=0; j<M; j++)                m[i][j]=0;    }};Matrix operator * (Matrix &a,Matrix &b){    Matrix c;    c.clearO();    for(int k=0; k<M; k++)        for(int i=0; i<M; i++)        {            if(a.m[i][k]==0) continue;             for(int j=0; j<M; j++)            {                if(b.m[k][j]==0) continue;                 c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j])%MOD;            }        }    return c;}Matrix operator ^ (Matrix &a,LL b){    Matrix c;    c.clearI();    while(b)    {        if(b&1) c=c*a;        b>>=1;        a=a*a;    }    return c;}int main(){    LL n,k,m;    while(~scanf("%I64d%I64d%I64d",&n,&k,&m))    {        MOD = m;        Matrix a,b;        a.clearO(),b.clearO();        for(int i=0; i<n; i++)        {            b.m[i][n+i]=b.m[n+i][n+i]=1;            for(int j=0; j<n; j++)            {                scanf("%I64d",&a.m[i][j]);                b.m[i][j]=a.m[i][j];            }        }        b=b^(k);        a=a*b;        for(int i=0; i<n; i++)        {            for(int j=0; j<n; j++)            {                if(j) printf(" ");                printf("%I64d",a.m[i][j+n]);            }            puts("");        }    }    return 0;}

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