等比数列二分求和取模

Posted 2021-03-18 nonames

题意：Given a n × n matrix A and a positive integer k, find the sum S = A + A² + A³ + … + A^k.

n (n ≤ 30), k (k ≤ 10⁹) and m (m < 10⁴)

输出结果矩阵

解法：

若 n是偶数

Sn= a+...+an/2 + an/2+1

+ an/2+2 +...+ an/2+n/2

=(a+...+an/2) + an/2(a+...+an/2)

=Sn/2+ an/2Sn/2

=(1+an/2)Sn/2等比数列二分求和取模

 

2) 若n是奇数

Sn= a+...+a(n-1)/2 + a(n-1)/2+1

+...

+ a(n-1)/2+(n-1)/2 + a(n-1)/2+(n-1)/2 + 1

=S(n-1)/2

+ a(n-1)/2(a+...+a(n-1)/2)+an

=(1+a(n-1)/2)S(n-1)/2+an

//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdio.h>
#include <queue>
#include <stack>;
#include <map>
#include <set>
#include <ctype.h>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f
#define PI acos(-1)
using namespace std;
typedef long long ll ;
int n , k , mod ;

struct node
{
    int a[49][49];
    node(){
        memset(a , 0 , sizeof(a));
    }
};

node M;

node mul(node A , node B)
{
    node C ;
    for(int i = 0 ; i < n ; i++)
    {
        for(int j = 0 ; j < n ; j++)
        {
            for(int k = 0 ; k < n ; k++)
            {
                C.a[i][j] = (C.a[i][j] + A.a[i][k] * B.a[k][j]) % mod ;
            }
        }
    }
    return C ;
}

node quickpow(node A , int t)
{
    node ans ;
    for(int i = 0 ; i < n ; i++)
    {
        ans.a[i][i] = 1 ;
    }
    while(t)
    {
        if(t&1)
        {
            ans = mul(ans , A) ;
        }
        t >>= 1 ;
        A = mul(A , A);
    }
    return ans ;
}

node jia(node A , node B)
{
    for(int i = 0 ; i < n ; i++)
    {
        for(int j = 0 ; j < n ; j++)
        {
            A.a[i][j] = (A.a[i][j] + B.a[i][j]) % mod ;
        }
    }
    return A ;
}

node half(node A , int k)
{
    if(k == 1)
        return A ;
    else if(k % 2 == 0)
    {
        return mul(half(A , k/2) , jia(quickpow(A , k/2) , M));
    }
    else{
        return jia(mul(half(A , (k-1)/2) , jia(quickpow(A , (k-1)/2) , M)) , quickpow(A , k));
    }
}

int main()
{
    /*#ifdef ONLINE_JUDGE
    #else
        freopen("D:/c++/in.txt", "r", stdin);
        freopen("D:/c++/out.txt", "w", stdout);
    #endif*/
    scanf("%d%d%d" , &n , &k , &mod);
    node A , B ;
    for(int i = 0 ; i < n ; i++)
    {
        M.a[i][i] = 1 ;
        for(int j = 0 ; j < n ; j++)
        {
            scanf("%d" , &A.a[i][j]);
        }
    }
    B = half(A , k);
    for(int i = 0 ; i < n ; i++)
    {
        for(int j = 0 ; j < n ; j++)
        {
            if(j == 0)
                cout << B.a[i][j] ;
            else{
                cout << " " << B.a[i][j] ;
            }
        }
        cout << endl ;
    }

    return 0 ;
}

解法二：等比矩阵

|A  E|

|0  E|

|A  ,  E|                　　 |A^n , 1+A^1+A^2+....+A^(n-1)|

|0  ,  E| 的n次方等于　|0     , 　　                  1         |

构造一个大矩阵，进行快速幂后，输出右上角矩阵。

//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdio.h>
#include <queue>
#include <stack>;
#include <map>
#include <set>
#include <ctype.h>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f
#define PI acos(-1)
using namespace std;
typedef long long ll ;
int n , k , mod ;

struct node
{
    int a[200][200];
    node(){
        memset(a , 0 , sizeof(a));
    }
};


node mul(node A , node B)
{
    node C ;
    for(int i = 0 ; i < n ; i++)
    {
        for(int j = 0 ; j < n ; j++)
        {
            for(int k = 0 ; k < n ; k++)
            {
                C.a[i][j] = (C.a[i][j] + A.a[i][k] * B.a[k][j]) % mod ;
            }
        }
    }
    return C ;
}

node quickpow(node A , int t)
{
    node ans ;
    for(int i = 0 ; i < n ; i++)
    {
        ans.a[i][i] = 1 ;
    }
    while(t)
    {
        if(t&1)
        {
            ans = mul(ans , A) ;
        }
        t >>= 1 ;
        A = mul(A , A);
    }
    return ans ;
}


int main()
{
    /*#ifdef ONLINE_JUDGE
    #else
        freopen("D:/c++/in.txt", "r", stdin);
        freopen("D:/c++/out.txt", "w", stdout);
    #endif*/
    node A , B;
    scanf("%d%d%d" , &n , &k , &mod);
    for(int i = 0 ; i < n ; i++)
    {
        for(int j = 0 ; j < n ; j++)
        {
            scanf("%d" , &A.a[i][j]);
            if(i == j)
            {
                A.a[i][j+n] = 1 ;
                A.a[i+n][j+n] = 1 ;
            }
        }
    }
    n *= 2 ;
    B = quickpow(A , k+1);
    for(int i = 0 ; i < n/2 ; i++)
    {
        for(int j = n/2 ; j < n ; j++)
        {
            if(i+n/2 == j)  B.a[i][j]--;
            if(j == n/2)
                cout << B.a[i][j]  ;
            else{
                cout << " " << B.a[i][j] ;
            }
        }
        cout << endl ;
    }

    return 0 ;
}