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一、学习资料

Blog1, Blog2,

论文1 -> 俞华程《矩阵乘法在信息学中的应用》

二、题目

1. hdoj-1575

描述：求矩阵A^k的主对角线元素和 (mod 9973).

PS: 入门题[二分求矩阵幂]

 

#include <stdio.h>
#include <string.h>
#define NL 10
#define MD 9973

struct Matrix {
int v[NL][NL];
Matrix () {
memset(v, 0, sizeof(v));
}
//矩阵相乘
void multiply(Matrix m, int n) {
Matrix m1;
for (int i=0; i<n; i++) {
for (int j=0; j<n; j++) {
for (int k=0; k<n; k++) {
m1.v[i][j] += v[i][k]*m.v[k][j];
if (m1.v[i][j] >= MD) m1.v[i][j] %= MD;
}
}
}
for (int i=0; i<n; i++) {
for (int j=0; j<n; j++)
v[i][j] = m1.v[i][j];
}
}
void print(int n) {
for (int i=0; i<n; i++) {
for (int j=0; j<n; j++)
printf("%d ", v[i][j]);
printf("\n");
}
}
};

int main()
{
// freopen("data.in", "r", stdin);
// freopen("data.out", "w", stdout);
int t;
int n, k, k0;
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &k);
Matrix m0, m1;
for (int i=0; i<n; i++) {
for (int j=0; j<n; j++)
scanf("%d", &m0.v[i][j]);
}
m1 = m0;
/*二分求矩阵幂*/
k--;
while (k>0) {
if (k&1) m1.multiply(m0,n);
m0.multiply(m0,n);
k = k>>1;
}
int ans = 0;
for (int i=0; i<n; i++) {
ans += m1.v[i][i];
ans %= MD;
}
printf("%d\n", ans);
}
return 0;
}

 

 

2. hdoj-1757

描述：

If x < 10， f(x) = x
If x >= 10， f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10)

构造递推矩阵：

 

#include <stdio.h>
#include <string.h>
#define NL 10
#define FOR(i,k,n) for(i=k;i<n;i++)

int md;
struct Matrix {
int v[NL][NL];
Matrix() {
memset(v, 0, sizeof(v));
}
void Multiply(Matrix m) {
int i, j, k;
Matrix m1;
FOR(i,0,NL) {
FOR(j,0,NL) {
FOR(k,0,NL) {
m1.v[i][j] += v[i][k]*m.v[k][j];
if (m1.v[i][j]>=md) m1.v[i][j] %= md;
}
}
}
FOR(i,0,NL) {
FOR(j,0,NL) {
v[i][j] = m1.v[i][j];
}
}
}
};

int main() {
int k;
while (scanf("%d%d", &k, &md) != EOF) {

Matrix m0, m1;
int i, j;
FOR(i,0,NL) {
// printf("ok!");
scanf("%d", &m0.v[0][i]);
}

if (k<10) {
printf("%d\n", k);
continue;
}

FOR(i,1,NL) {
m0.v[i][i-1] = 1;
}
k -= 10;
m1 = m0;
while (k>0) {
if (k&1) m0.Multiply(m1);
m1.Multiply(m1);
k >>= 1;
}
int ans = 0;
FOR(i,0,NL) {
ans += m0.v[0][i]*(NL-i-1);
if (ans>=md) ans %= md;
}
printf("%d\n", ans);
}
}

 

 

 

3. hdoj-2256

描述：求

分析：

 

#include <stdio.h>
#include <string.h>
#define NL 4
#define MD 1024
#define FOR(i,k,n) for(i=k;i<n;i++)

struct Matrix {
int v[NL][NL];
Matrix() {
memset(v, 0, sizeof(v));
}
void Multiply(Matrix m) {
int i, j, k;
Matrix m1;
FOR(i,0,NL) {
FOR(j,0,NL) {
FOR(k,0,NL) {
m1.v[i][j] += v[i][k]*m.v[k][j];
if (m1.v[i][j]>=MD) m1.v[i][j] %= MD;
}
}
}
FOR(i,0,NL) {
FOR(j,0,NL) {
v[i][j] = m1.v[i][j];
}
}
}
};

int main() {
int t, n;
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
Matrix m0, m1;
m0.v[0][0] = 5;
m0.v[0][1] = 12;
m0.v[1][0] = 2;
m0.v[1][1] = 5;
m1 = m0;
n--;
while (n>0) {
if (n&1) m0.Multiply(m1);
m1.Multiply(m1);
n >>= 1;
}
int ans = m0.v[0][0];
ans = (ans*2-1+MD)%MD;
printf("%d\n", ans);
}
return 0;
}

 

4.poj-3233

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

分析:

  1. if k is odd then S = A + ... + A^k/2 +  A^k/2+1(A + ... + A^k/2) ,else   S = A + ... + A^k/2 +  A^k/2(A + ... + A^k/2),

2. goto calculate k/2.

ps:用c++写的，效率很低，有待改进。

#include <stdio.h>
#include <string.h>
#define NL 30

int n, md;

class Matrix {
public:
int v[NL][NL];

Matrix() {
memset(v, 0, sizeof (v));
}

Matrix & operator =(const Matrix &m) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++)
this->v[i][j] = m.v[i][j];
}
return *this;
}

Matrix Multiply(Matrix m) {
Matrix m1;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
for (int k = 0; k < n; k++) {
m1.v[i][j] += v[i][k] * m.v[k][j];
m1.v[i][j] %= md;
}
}
}
return m1;
}

Matrix Add(Matrix m) {
Matrix m1;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
m1.v[i][j] = (v[i][j] + m.v[i][j]) % md;
}
}
return m1;
}

Matrix Pow(int k) {
Matrix m1, m0;
m1 = m0 = *this;
k--;
while (k > 0) {
if (k & 1) m1 = m1.Multiply(m0);
m0 = m0.Multiply(m0);
k >>= 1;
}
return m1;
}

void Print() {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++)
printf("%d ", v[i][j]);
printf("\n");
}
}
};
Matrix m0, m1;

Matrix cal(int k) {
Matrix m;
if (k == 1) return m0;
if (k & 1) {
Matrix mx = cal(k / 2);
Matrix my = m0.Pow(k / 2 + 1);
return mx.Add(mx.Multiply(my).Add(my));
} else {
Matrix mx = cal(k / 2);
Matrix my = m0.Pow(k / 2);
return mx.Add(my.Multiply(mx));
}
}

int main() {
int k;
scanf("%d%d%d", &n, &k, &md);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++)
scanf("%d", &m0.v[i][j]), m0.v[i][j] %= md;
}
m1 = cal(k);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
printf("%d%c", m1.v[i][j], (j == n - 1) ? '\n' : ' ');
}
}
return 0;
}