等比数列二分求和取模
题意：Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.
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 ;
}
