Matrix Power Series
Time Limit:
3000MS
Memory Limit: 131072K
Description
Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.
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
题意要求矩阵S=A+A^2+A^3+...+A^k mod m,可以用二分的方法。
单个数的快速幂请参考这篇文章。
快速等比数列求和:如何求k=19的情况呢?首先根据19的二进制10011构造这么一阶差分序列:1,2,16,所以原序列是0,1,3,19.
快速等比数列求和:如何求k=19的情况呢?首先根据19的二进制10011构造这么一阶差分序列:1,2,16,所以原序列是0,1,3,19.
那怎么求A+A^2+A^3+...+A^19呢?我们就按照原序列来求,也就要先求一阶差分序列。
由于第一个数是0,我们初始化ans=0
第二个数是1,ans=ans*A^1+A=A
第三个数是3,ans=ans*A^2+A(I+A)=A+A^2+A^3
由于二进制中间有两个0,所以与ans相乘的数变为A^16(算法:A^2*A^2=A^4,A^4*A^4=A^16)
与ans*A^16相加的数变为A+...A^16(算法:(A+A^2)*((I+A^2)=A+...+A^4,(A+...+A^4)*(I+A^4)=A+...+A^8,(A+...+A^8)*(I+A^8)=A+...A^16)
第四个数是19,ans=ans*A^16+(A+...A^16)
完整代码:
/*79ms,144KB*/
#include <cstdio>
#include <cstring>
const int matSize = 31;
int calSize = matSize, mod = 1000000007;
struct Matrix
{
int val[matSize][matSize];
Matrix(bool Init = false)///使用默认实参以减少代码量
{
for (int i = 0; i < calSize; i++)
{
for (int j = 0; j < calSize; j++)
val[i][j] = 0;
if (Init)
val[i][i] = 1;///单位矩阵
}
}
void print()
{
for (int i = 0; i < calSize; i++)
{
for (int j = 0; j < calSize; j++)
{
if (j)
putchar(' ');
printf("%d", val[i][j]);
}
puts("");
}
}
} Base;
Matrix operator + (Matrix &_a, Matrix &_b)
{
Matrix ret;
for (int i = 0; i < calSize; i++)
for (int j = 0; j < calSize; j++)
ret.val[i][j] = (_a.val[i][j] + _b.val[i][j]) % mod;
return ret;
}
Matrix operator * (Matrix &_a, Matrix &_b)
{
Matrix ret = Matrix();
for (int i = 0; i < calSize; i++)
for (int k = 0; k < calSize; k++)
if (_a.val[i][k])///优化一下
for (int j = 0; j < calSize; j++)
{
ret.val[i][j] += _a.val[i][k] * _b.val[k][j];
ret.val[i][j] %= mod;
}
return ret;
}
void deal(int k)
{
Matrix one = Matrix(true), ans = Matrix();
Matrix ep = Base, tmp = Base, temp;
while (k)
{
if (k & 1)
{
ans = ans * ep;
ans = ans + tmp;
}
temp = one + ep;
tmp = tmp * temp;
ep = ep * ep;
k >>= 1;
}
ans.print();
}
int main()
{
int k;
while (~scanf("%d%d%d", &calSize, &k, &mod))
{
for (int i = 0; i < calSize; i++)
for (int j = 0; j < calSize; j++)
{
scanf("%d", &Base.val[i][j]);
Base.val[i][j] %= mod;
}
deal(k);
}
return 0;
}