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[30][30];
    Matrix(){
        memset(m, 0, sizeof(m));
    }
}Matrix;
Matrix mtAdd(Matrix A, Matrix B){
    for(int i = 0;i < n;i ++)
        for(int j = 0;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 = 0;i < n;i ++)
        for(int j = 0;j < n;j ++)
            for(int k = 0;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 == 1) return A;
    Matrix tmp = mtPow(A, k >> 1);
    Matrix res = mtMul(tmp, tmp);
    if(k&1) res = mtMul(res, A);
    return res;
}
Matrix mtSum(Matrix A, int k){
    if(k == 1) return A;
    Matrix tmp = mtSum(A, k/2);
    if(k&1){
        Matrix t = mtPow(A, k/2+1);
        Matrix tmp1 = mtMul(tmp, t);
        Matrix tmp2 = mtAdd(t, tmp);
        return mtAdd(tmp1, tmp2);
    }else return mtAdd(tmp, mtMul(mtPow(A, k/2), tmp));
}
int main(){
    int k, tmp;
    /* freopen("in.c", "r", stdin); */
    while(~scanf("%d%d%d", &n, &k, &m)){
        Matrix M;
        for(int i = 0;i < n;i ++)
            for(int j = 0;j < n;j ++){
                scanf("%d", &tmp);
                M.m[i][j] = tmp;
            }
        M = mtSum(M, k);
        for(int i = 0;i < n;i ++){
            for(int j = 0;j < n;j ++)
                printf("%d ", M.m[i][j]);
            puts("");
        }
    }
    return 0;
}