我正在进行一项小型python挑战，其中代码在我的笔记本电脑上工作正常，但在在线控制台的第18行显示索引错误。 问题是
Doomsday Fuel
=============
Making fuel for the LAMBCHOP's reactor core is a tricky process because of the exotic matter involved. It starts as raw ore, then during processing, begins randomly changing between forms, eventually reaching a stable form. There may be multiple stable forms that a sample could ultimately reach, not all of which are useful as fuel.
Commander Lambda has tasked you to help the scientists increase fuel creation efficiency by predicting the end state of a given ore sample. You have carefully studied the different structures that the ore can take and which transitions it undergoes. It appears that, while random, the probability of each structure transforming is fixed. That is, each time the ore is in 1 state, it has the same probabilities of entering the next state (which might be the same state). You have recorded the observed transitions in a matrix. The others in the lab have hypothesized more exotic forms that the ore can become, but you haven't seen all of them.
Write a function answer(m) that takes an array of array of nonnegative ints representing how many times that state has gone to the next state and return an array of ints for each terminal state giving the exact probabilities of each terminal state, represented as the numerator for each state, then the denominator for all of them at the end and in simplest form. The matrix is at most 10 by 10. It is guaranteed that no matter which state the ore is in, there is a path from that state to a terminal state. That is, the processing will always eventually end in a stable state. The ore starts in state 0. The denominator will fit within a signed 32-bit integer during the calculation, as long as the fraction is simplified regularly.
For example, consider the matrix m:
[
[0,1,0,0,0,1], # s0, the initial state, goes to s1 and s5 with equal probability
[4,0,0,3,2,0], # s1 can become s0, s3, or s4, but with different probabilities
[0,0,0,0,0,0], # s2 is terminal, and unreachable (never observed in practice)
[0,0,0,0,0,0], # s3 is terminal
[0,0,0,0,0,0], # s4 is terminal
[0,0,0,0,0,0], # s5 is terminal
]
So, we can consider different paths to terminal states, such as:
s0 -> s1 -> s3
s0 -> s1 -> s0 -> s1 -> s0 -> s1 -> s4
s0 -> s1 -> s0 -> s5
Tracing the probabilities of each, we find that
s2 has probability 0
s3 has probability 3/14
s4 has probability 1/7
s5 has probability 9/14
So, putting that together, and making a common denominator, gives an answer in the form of
[s2.numerator, s3.numerator, s4.numerator, s5.numerator, denominator] which is
[0, 3, 2, 9, 14].
Languages
=========
To provide a Python solution, edit solution.py
To provide a Java solution, edit solution.java
Test cases
==========
Inputs:
(int) m = [[0, 2, 1, 0, 0], [0, 0, 0, 3, 4], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]
Output:
(int list) [7, 6, 8, 21]
Inputs:
(int) m = [[0, 1, 0, 0, 0, 1], [4, 0, 0, 3, 2, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]
Output:
(int list) [0, 3, 2, 9, 14]
代码是
from itertools import product
from fractions import Fraction
from functools import reduce
#this is for matrix inversion
def invert(matrix):
n = len(matrix)
inverse = [[Fraction(0) for col in range(n)] for row in range(n)]
for i in range(n):
inverse[i][i] = Fraction(1)
for i in range(n):
for j in range(n):
if i != j:
if matrix[i][i] == 0:
return False
ratio = matrix[j][i] / matrix[i][i]
for k in range(n):
inverse[j][k] = inverse[j][k] - ratio * inverse[i][k]
matrix[j][k] = matrix[j][k] - ratio * matrix[i][k]
for i in range(n):
a = matrix[i][i]
if a == 0:
return False
for j in range(n):
inverse[i][j] = inverse[i][j] / a
return inverse
#finding sum of a row in matrix
def sumRow(m, r):
return sum(m[r])
#subtracting two matrices
def substract(matr_a, matr_b):
output = []
for i in range(len(matr_a)):
tmp = []
for valA, valB in zip(matr_a[i], matr_b[i]):
tmp.append(valA - valB)
output.append(tmp[:])
return output[:]
#matrix multiplication
def matrixmult(matr_a, matr_b):
#cols = len(matr_b[0])
rows = len(matr_b)
if rows is not 0:
cols = len(matr_b[0])
else:
cols = 0
resRows = range(len(matr_a))
rMatrix = [[0] * cols for _ in resRows]
for idx in resRows:
for j, k in product(range(cols), range(rows)):
rMatrix[idx][j] += matr_a[idx][k] * matr_b[k][j]
if cols is not 0:
return rMatrix
else:
return 0
# return rMatrix
#gcd to find lcm
def gcd(a, b):
while b:
a, b = b, a % b
return a
#lcm to find last value of the output
def lcm(a,n):
ans = a[0]
for i in range(1,n):
ans = (a[i]*ans)//gcd(a[i],ans)
return ans
#main function
def answer(m):
num = len(m)
f=[]
#finding zero rows
for i in range(0,num):
j=[]
j.append(sumRow(m,i))
j.append(i)
f.append(j)
k=0;
#Fraction Conversion
for i in range(0,num):
for j in range(0,num):
if f[i][0]!=0:
m[i][j]=Fraction(m[i][j],f[i][0])
j=[]
for i in range(0,len(f)):
if f[i][0].numerator==0:
j.append(i)
del m[f[i][1]-k]
k=k+1
q=m
k=[]
t=0
e=m
# print(m)
q=[]
for i in range(0,len(m)):
row=[]
for r in range(0,len(m[0])):
for t in range(0,len(j)):
if j[t] is r:
row.append(m[i][r])
q.append(row)
# print(q)
t=0;
w=0;
e=[]
for i in range(0,len(m)):
row=[]
flag=1
for r in range(0,len(m[0])):
for t in range(0,len(j)):
if j[t] is r:
flag=0
break
if flag is 1:
row.append(m[i][r])
e.append(row)
#print(e)
l=[]
for i in range(0,len(e)):
k=[]
for b in range(0,len(e)):
if i==b:
k.append(1)
else:
k.append(0)
l.append(k)
#print(l)
#print(e)
l=substract(l,e)
#print(l)
l=invert(l)
r = matrixmult(l,q)
#print(r)
if r == 0:
return 0
else:
m =r[0]
e=[]
for i in range(0,len(m)):
e.append(m[i].denominator)
k=lcm(e,len(e))
e=[]
for i in range(0,len(m)):
e.append((m[i].numerator*k)//m[i].denominator)
e.append(k)
#print(e)
return e
#answer (m = [[0, 2, 1, 0, 0], [0, 0, 0, 3, 4], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]])
#answer (m = [[0, 1, 0, 0, 0, 1], [4, 0, 0, 3, 2, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]])
剩下的时间大概是19h请告诉我我在哪里做错了我是python的新手刚刚开始在python中编码，因为与java相比，处理分数很容易。
注意：在线控制台不接受numpy库所以我们必须硬编码像矩阵反转这样的东西
在测试各种情况后，我发现了这个错误
Traceback (most recent call last):
File "C:\Users\python.py", line 148, in <module>
answer (m = [[0, 2, 1, 0, 0], [0, 0, 0, 3, 4], [0, 0, 0, 0, 0], [0, 4, 0, 0, 0], [0, 0, 0, 0, 0]])
File "C:\Users\python.py", line 131, in answer
l=invert(l)
File "C:\Users\python.py", line 18, in invert
matrix[j][k] = matrix[j][k] - ratio * matrix[i][k]
IndexError: list index out of range
答案 0 :(得分：0)
当您为行矩阵[j] [k] =矩阵[j] [k] - 比率*矩阵[i] [k]引用的列表索引不在列表中时，会生成该错误。我不太确定为什么它会生成，因为我测试了def反向（矩阵）的代码并且它通过了我的所有测试用例。你明白了吗？