题目https://www.dotcpp.com/oj/problem1472.html
1 普通做法 循环嵌套
n,m = list(map(int,input().split()))
mat=[]
for i in range(n):
row=list(map(int,input().split()))
mat.append(row)
C=mat.copy()#mat[:]
def cf(A,B):
x = [[0]*n for i in range(n)]
for i in range(len(A)):
for j in range(len(B[0])):
for k in range(len(B)):
x[i][j] += A[i][k]*B[k][j]
return x
for i in range(m-1):
mat=cf(mat,C)
for i in range(n):
for j in range(n):
print(mat[i][j],end=" ")
print()
#还应该考虑特殊情况 0次幂 特殊用例
2.zip(*A)可以把矩阵横纵坐标交换,即转置
矩阵乘法->以两个二阶矩阵举例,即第一个矩阵的第一行数据分别乘以对应的第二个矩阵的第一列数据,思路就放在如何将不同矩阵的行与列的数值之间在数组中的索引进行统一上
先将要求的矩阵转置,然后只有按位置一一对应相乘就行,后面再将得到的答案转换二维列表输出即可
def f(A,n,m):
if m == 0:
C = [[1 if i == j else 0 for j in range(n)] for i in range(n)]
else:
C = A[:]
for i in range(m-1):
C = [[sum(a*b for a,b in zip(row,col)) for col in zip(*A)] for row in C]
for i in range(n):
for j in range(n):
print(C[i][j],end=' ')
print()
if __name__ == '__main__':
n,m = map(int,input().strip().split())
A = [[int(j) for j in input().strip().split()] for i in range(n)]
f(A,n,m)
3.快速幂 ,真正考点!
https://blog.dotcpp.com/a/77575 题解
快速幂模版
def fastpow(base,n)
ans=1
while(n):
if n&1:
ans*=base
base*=base
n>>=1
return ans
def fpowx(x, n):
res = 1
while n:
if n & 1:
res = res * x
# compute x^2 x^4 x^8
x *= x
n >>= 1
return res
矩阵快速幂
#先定义矩阵的乘法
def multi(A,B):
x = [[0]*n for i in range(n)]
for i in range(len(A)):
for j in range(len(B[0])):
for k in range(len(B)):
x[i][j] += A[i][k]*B[k][j]
return x
#计算快速幂
def quick_multi(matrix,m):
res=[[1 if i == j else 0 for j in range(n)] for i in range(n)]
while m:
if m&1:
res=multi(res,matrix)
matrix=multi(matrix,matrix)
m=m>>1
return res