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

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;
}