POJ -- 3233 求“等比矩阵”前n（n <=10^9）项和

Matrix Power Series

 

Description

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

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

思路：1.最基本的，需要用到矩阵快速幂 2.快速幂求完之后怎样快速求和？若逐项累加求和必然会超时，这时需要求递推公式：（1）若n为偶数，则：S(n) = A^(n/2)*S(n/2)+s(n/2);(2)若n为奇数 S(n) = A^(n/2+1) + S(n/2)*A^(n/2+1) + S(n/2),公式不难推，写几个就发现规律了。这样就把时间复杂度降下来了。

 #include<cstdio>
 #include<string>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 int n, m;
 typedef struct Matrix{
     int m[][];
     Matrix(){
         memset(m, , sizeof(m));
     }
 }Matrix;
 Matrix mtAdd(Matrix A, Matrix B){
     for(int i = ;i < n;i ++)
         for(int j = ;j < n;j ++){
             A.m[i][j] += B.m[i][j];
             A.m[i][j] %= m;
         }
     return A;
 }
 Matrix mtMul(Matrix A, Matrix B){
     Matrix tmp;
     for(int i = ;i < n;i ++)
         for(int j = ;j < n;j ++)
             for(int k = ;k < n;k ++){
                 tmp.m[i][j] += A.m[i][k]*B.m[k][j];
                 tmp.m[i][j] %= m;
             }
     return tmp;
 }
 Matrix mtPow(Matrix A, int k){
     if(k == ) return A;
     Matrix tmp = mtPow(A, k >> );
     Matrix res = mtMul(tmp, tmp);
     if(k&) res = mtMul(res, A);
     return res;
 }
 Matrix mtSum(Matrix A, int k){
     if(k == ) return A;
     Matrix tmp = mtSum(A, k/);
     if(k&){
         Matrix t = mtPow(A, k/+);
         Matrix tmp1 = mtMul(tmp, t);
         Matrix tmp2 = mtAdd(t, tmp);
         return mtAdd(tmp1, tmp2);
     }else return mtAdd(tmp, mtMul(mtPow(A, k/), tmp));
 }
 int main(){
     int k, tmp;
     /* freopen("in.c", "r", stdin); */
     while(~scanf("%d%d%d", &n, &k, &m)){
         Matrix M;
         for(int i = ;i < n;i ++)
             for(int j = ;j < n;j ++){
                 scanf("%d", &tmp);
                 M.m[i][j] = tmp;
             }
         M = mtSum(M, k);
         for(int i = ;i < n;i ++){
             for(int j = ;j < n;j ++)
                 printf("%d ", M.m[i][j]);
             puts("");
         }
     }
     return ;
 }