poj3233(81/600)

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

等比矩陣這玩意…
a1
01
n次方以後…
a^n 1+…+a^n-1
0 1

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
int n,k,mo;
struct p
{
    int q[65][65];
};
p cheng(p q,p w,int t)
{
    p fs;
    memset(fs.q,0,sizeof(fs.q));
    for(int a=1;a<=t;a++)
    {
        for(int b=1;b<=t;b++)
        {
            for(int c=1;c<=t;c++)
            {
                fs.q[a][b]+=q.q[a][c]*w.q[c][b]%mo;
                fs.q[a][b]%=mo;
            }
        }
    }
    return fs;
}
p ksm(p ds,int zs,int t)
{
    p fs;
    for(int a=1;a<=t;a++)
    {
        for(int b=1;b<=t;b++)
        {
            if(a==b)fs.q[a][b]=1;
            else fs.q[a][b]=0;
        }
    }
    while(zs)
    {
        if(zs&1)fs=cheng(fs,ds,t);
        ds=cheng(ds,ds,t);
        zs>>=1;
    }
    return fs;
}
p ys;
int main()
{
    cin>>n>>k>>mo;
    for(int a=1;a<=n;a++)
    {
        for(int b=1;b<=n;b++)
        {
            scanf("%d",&ys.q[a][b]);
            if(a==b)ys.q[a][b+n]=ys.q[a+n][b+n]=1;
        }
    }
    ys=ksm(ys,k+1,2*n);
    for(int a=1;a<=n;a++)
    {
        for(int b=1;b<=n;b++)
        {
            if(a!=b)continue;
            ys.q[a][b+n]--;
            if(ys.q[a][b+n]>=0)continue;
            ys.q[a][b+n]+=mo;
        }
    }
    for(int a=1;a<=n;a++)
    {
        for(int b=1;b<=n;b++)
        {
            cout<<ys.q[a][b+n];
            if(b<n)cout<<" ";
        }
        cout<<endl;
    }
}