UVA - 11149 Power of Matrix（矩阵倍增）

题意：已知N*N的矩阵A，输出矩阵A + A² + A³ + . . . + A^k，每个元素只输出最后一个数字。

分析：

A + A² + A³ + . . . + Aⁿ可整理为下式，

从而可以用log₂(n)的复杂度算出结果。

注意：输入时把矩阵A的每个元素对10取余，因为若不处理，会导致k为1的时候结果出错。

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cctype>

#include<cmath>

#include<iostream>

#include<sstream>

#include<iterator>

#include<algorithm>

#include<string>

#include<vector>

#include<set>

#include<map>

#include<stack>

#include<deque>

#include<queue>

#include<list>

#define lowbit(x) (x & (-x))

const double eps = 1e-8;

inline int dcmp(double a, double b){

    if(fabs(a - b) < eps) return 0;

    return a > b ? 1 : -1;

}

typedef long long LL;

typedef unsigned long long ULL;

const int INT_INF = 0x3f3f3f3f;

const int INT_M_INF = 0x7f7f7f7f;

const LL LL_INF = 0x3f3f3f3f3f3f3f3f;

const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;

const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};

const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};

const int MOD = 10;

const double pi = acos(-1.0);

const int MAXN = 40 + 10;

const int MAXT = 10000 + 10;

using namespace std;

int n;

struct Matrix{

    int r, c, matrix[MAXN][MAXN];

    Matrix(int rr, int cc):r(rr), c(cc){

        memset(matrix, 0, sizeof matrix);

    }

};

Matrix add(Matrix a, Matrix b){

    Matrix ans(n, n);

    for(int i = 0; i < a.r; ++i){

        for(int j = 0; j < a.c; ++j){

            ans.matrix[i][j] = ((a.matrix[i][j] % MOD) + (b.matrix[i][j] % MOD)) % MOD;

        }

    }

    return ans;

}

Matrix mul(Matrix a, Matrix b){

    Matrix ans(a.r, b.c);

    for(int i = 0; i < a.r; ++i){

        for(int j = 0; j < b.c; ++j){

            for(int k = 0; k < a.c; ++k){

                (ans.matrix[i][j] += ((a.matrix[i][k] % MOD) * (b.matrix[k][j] % MOD)) % MOD) %= MOD;

            }

        }

    }

    return ans;

}

Matrix QPOW(Matrix a, int k){

    Matrix ans(n, n);

    for(int i = 0; i < n; ++i){

        ans.matrix[i][i] = 1;

    }

    while(k){

        if(k & 1) ans = mul(ans, a);

        a = mul(a, a);

        k >>= 1;

    }

    return ans;

}

Matrix solve(Matrix tmp, int k){

    if(k == 1) return tmp;

    Matrix t = solve(tmp, k >> 1);

    Matrix ans = add(t, mul(QPOW(tmp, k >> 1), t));

    if(k & 1) ans = add(ans, QPOW(tmp, k));

    return ans;

}

int main(){

    int k;

    while(scanf("%d%d", &n, &k) == 2){

        if(n == 0) return 0;

        Matrix tmp(n, n);

        for(int i = 0; i < n; ++i){

            for(int j = 0; j < n; ++j){

                scanf("%d", &tmp.matrix[i][j]);

                tmp.matrix[i][j] %= MOD;

            }

        }

        Matrix ans = solve(tmp, k);

        for(int i = 0; i < ans.r; ++i){

            for(int j = 0; j < ans.c; ++j){

                if(j) printf(" ");

                printf("%d", ans.matrix[i][j]);

            }

            printf("\n");

        }

        printf("\n");

    }

    return 0;

}

巴特西

UVA - 11149 Power of Matrix（矩阵倍增）

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