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Matrix Power Series POJ - 3233 (矩陣快速冪)

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題意:

題解:

附上程式碼:


#include<iostream>
#include<cstdio>

using namespace std;

typedef long long ll;

const int MAXN=70;

struct node{
    ll a[MAXN][MAXN];
};
node shu,ans,mp;

ll N;
ll n,kk,mod;

node matrix(node x,node y)
{
    for(int i=1;i<=N;i++){
        for(int j=1;j<=N;j++){
            mp.a[i][j]=0;
            for(int p=1;p<=N;p++){
                mp.a[i][j]=(mp.a[i][j]+x.a[i][p]*y.a[p][j]+mod)%mod;
            }
        }
    }
    return mp;
}

void work(ll k)
{
    for(int i=1;i<=N;i++){
        for(int j=1;j<=N;j++){
            ans.a[i][j]=0;
        }
    }
    for(int i=1;i<=N;i++){
        ans.a[i][i]=1;
    }
    node t=shu;
    while(k){
        if(k&1){
            ans=matrix(ans,t);
        }
        k>>=1;
        t=matrix(t,t);
    }
}

int main()
{
    scanf("%lld%lld%lld",&n,&kk,&mod);
    N=2*n;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            scanf("%lld",&shu.a[i][j]);
        }
        shu.a[n+i][i]=shu.a[n+i][n+i]=1;
    }
    work(kk+1);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            ll a=ans.a[n+i][j]%mod;
            if(i==j){
                a=(a+mod-1)%mod;
            }
            printf("%lld%c",a,j==n?'\n':' ');
        }
    }
    return 0;
}