I am practicing 'Round D 2020 - Kick Start 2020' though I have properly done coding for Record Breaker problem and also showing correct answers for sample test cases still I am getting RE (Runtime error).
See attached screenshot and code.
enter image description here
T=int(input())
rlst=[]
if "\n" in T:
T
for i in range(T):
N=int(input())
V=input()
V_lst=V.split()
rbday=0
high_day=V_lst[0]
if V_lst[0] > V_lst[1]:
rbday=rbday+1
high_day=V_lst[0]
for n in range(1,N-1):
if ((V_lst[n] > V_lst[n-1]) and (V_lst[n] > V_lst[n+1]) and (V_lst[n] > high_day)) :
high_day=V_lst[n]
rbday=rbday+1
rlst.append(rbday)
for i in range(T):
print("Case {}# {}".format(i+1,rlst[i]))
The error is in this line: if V_lst[0] > V_lst[1]:, because you didn't consider the case where N is equal to 1, so this V_lst[1] is out of index.
I can share my solution, using Python 3.7 too:
T = int(input())
for case in range(1, T+1):
N = int(input())
V = list(map(int, input().split()))
n_breaks = 0
d = 0 #It's like previous high day in your code.
for i in range(N):
if N == 1:
n_breaks = 1
break
if i != N-1:
if V[i] > d and V[i] > V[i+1]:
n_breaks += 1
d = V[i]
else:
if V[i] > d and V[i] > V[i-1]:
n_breaks += 1
print(f"Case #{case}: {n_breaks}")
Related
I'm learning SageMath (uses Python 3)
and playing with the Goldbach conjecture.
I wrote this function (it works!):
def Goldbach(n):
if n % 2 != 0 or n <= 2:
show("No és parell")
else:
for i in srange(n):
if is_prime(i):
for j in srange(n):
if is_prime(j) and i + j == n:
a = [i, j]
show(a)
return
Now I'm trying (no idea) to do the following plot:
Denoting by r(2k) the number of Goldbach partitions of 2k,
the conjecture affirms that r(2k) > 0 if k > 1.
I have to do a graph of points (k, r(2k)), k > 2.
How could I do it?
First of all, let us get some better implementation in Sage
of the routine counting the number r(K) (for K > 2 some even integer) of the solutions for p + q = 2k, p, q prime numbers.
We count both solutions (p, q) and (q, p) when they differ.
def r(K):
if K not in ZZ or K <= 2 or K % 2:
return None
if K == 4:
return 1
count = 0
for p in primes(3, K):
for q in primes(3, K + 1 - p):
if p + q == K:
count += 1
return count
goldbach_points = [(K, r(K)) for K in range(4, 100,2)]
show(points(goldbach_points))
This gives:
I'm trying to translate a C routine from an old sound synthesis program into R, but have indexing issues which I'm struggling to understand (I'm a beginner when it comes to using loops).
The routine creates an exponential lookup table - the vector exptab:
# Define parameters
sinetabsize <- 8192
prop <- 0.8
BP <- 10
BD <- -5
BA <- -1
# Create output vector
exptab <- vector("double", sinetabsize)
# Loop
while(abs(BD) > 0.00001){
BY = (exp(BP) -1) / (exp(BP*prop)-1)
if (BY > 2){
BS = -1
}
else{
BS = 1
}
if (BA != BS){
BD = BD * -0.5
BA = BS
BP = BP + BD
}
if (BP <= 0){
BP = 0.001
}
BQ = 1 / (exp(BP) - 1)
incr = 1 / sinetabsize
x = 0
stabsize = sinetabsize + 1
for (i in (1:(stabsize-1))){
x = x + incr
exptab [[sinetabsize-i]] = 1 - (BQ * (exp(BP * x) - 1))
}
}
Running the code gives the error:
Error in exptab[[sinetabsize - i]] <- 1 - (BQ * (exp(BP * x) - 1)) :
attempt to select less than one element in integerOneIndex
Which, I understand from looking at other posts, indicates an indexing problem. But, I'm finding it difficult to work out the exact issue.
I suspect the error may lie in my translation. The original C code for the last few lines is:
for (i=1; i < stabsize;i++){
x += incr;
exptab[sinetabsize-i] = 1.0 - (float) (BQ*(exp(BP*x) - 1.0));
}
I had thought the R code for (i in (1:(stabsize-1))) was equivalent to the C code for (i=1; i< stabsize;i++) (i.e. the initial value of i is i = 1, the test is whether i < stabsize, and the increment is +1). But now I'm not so sure.
Any suggestions as to where I'm going wrong would be greatly appreciated!
As you say, array indexing in R starts at 1. In C it starts at zero. I reckon that's your problem. Can sinetabsize-i ever get to zero?
t=int(input())
while t>0 :
c=0
n,h,y1,y2,e = list(map(int, input().split()))
for i in range(n):
x0,x1 = list(map(int, input().split()))
if x0==1 :
if x1 < h-y1:
e -= 1
else :
if y2 < x1 :
e -= 1
if e>0 :
c+=1
else :
break
print(c)
t-=1
It is passing the sample test cases but on submission, it is showing runtime error(NZEC) occurred.
Here is the link to the question: https://www.codechef.com/problems/PIPSQUIK
The problem is that you're reading the inputs and processing them simultaneously. So, a situation can arise in some test cases such that e<=0 but you still have some x0 x1 to read(i.e. i<n-1). In such cases, you'll break the loop because e<=0 and in next iteration of while loop, you'll try to read 5 values n,h,y1,y2,e = list(map(int, input().split())) but you'll receive only 2 values x0 x1 and hence it'll throw a ValueError: not enough values to unpack (expected 5, got 2) and hence it'll not pass all the test cases.
To fix this, just take all inputs first and then process them according to your current logic.
t=int(input())
while t>0 :
c=0
n,h,y1,y2,e = list(map(int, input().split()))
inputs = []
for i in range(n):
inputs.append(list(map(int, input().split())))
for inp in inputs:
x0,x1 = inp
if x0==1 :
if x1 < h-y1:
e -= 1
else :
if y2 < x1 :
e -= 1
if e>0 :
c+=1
else :
break
print(c)
t -= 1
So here is my situation. Ive been trying to make a advanced calculator in python 3.4, one where you can just type something like this. '1 + 1', and it would then give you the answer of '2'. Now i will explain how my calculator is supposed to work. So you start by entering a maths equation, then it counts the words you entered based on the spaces. It does this so it knows how long some future loops need to be. Then it splits up everything that you entered. It splits it up into str's and int's but its all still in the same variable and it's all still in order. The thing i'm having trouble with is when it is meant to actually do the calculations.
here is all of my code-
# This is the part were they enter the maths equation
print("-------------------------")
print("Enter the maths equation")
user_input = input("Equation: ")
# This is were it counts all of the words
data_count = user_input.split(" ")
count = data_count.__len__()
# Here is were is splits it into str's and int's
n1 = 0
data = []
if n1 <= count:
for x in user_input.split():
try:
data.append(int(x))
except ValueError:
data.append(x)
n1 += 1
# And this is were it actually calculates everything
number1 = 0
number2 = 0
n1 = 0
x = 0
answer = 0
while n1 <= count:
#The code below checks if it is a number
if data[n1] < 0 or data[n1] > 0:
if x == 0:
number1 = data[n1]
elif x == 1:
number2 = data[n1]
elif data[n1] is "+":
if x == 0:
answer += number1
elif x == 1:
answer += number2
n1 += 1
x += 1
if x > 1:
x = 0
print("Answer =", answer)
but during the calculation it messes up and gives me and error
error-
if data[n1] < 0 or data[n1] > 0:
TypeError: unorderable types: str() < int()
can anyone see what i am doing wrong here?
Thanks
When you are comparing a string and an integer, this problem comes.
Python doesn't guess, it throws an error.
To fix this, simply call int() to convert your string to an integer:
int(input(...))
So, corrected statement should be:
if int(data[n1]) < 0 or int(data[n1]) > 0:
How can the Kendall tau distance (a.k.a. bubble-sort distance) between two permutations be calculated in R without loading additional libraries?
Here is an O(n.log(n)) implementation scraped together after reading around, but I suspect there may be better R solutions.
inversionNumber <- function(x){
mergeSort <- function(x){
if(length(x) == 1){
inv <- 0
#printind(' base case')
} else {
n <- length(x)
n1 <- ceiling(n/2)
n2 <- n-n1
y1 <- mergeSort(x[1:n1])
y2 <- mergeSort(x[n1+1:n2])
inv <- y1$inversions + y2$inversions
x1 <- y1$sortedVector
x2 <- y2$sortedVector
i1 <- 1
i2 <- 1
while(i1+i2 <= n1+n2+1){
if(i2 > n2 || (i1 <= n1 && x1[i1] <= x2[i2])){ # ***
x[i1+i2-1] <- x1[i1]
i1 <- i1 + 1
} else {
inv <- inv + n1 + 1 - i1
x[i1+i2-1] <- x2[i2]
i2 <- i2 + 1
}
}
}
return (list(inversions=inv,sortedVector=x))
}
r <- mergeSort(x)
return (r$inversions)
}
.
kendallTauDistance <- function(x,y){
return(inversionNumber(order(x)[rank(y)]))
}
If one needs custom tie-breaking one would have to fiddle with the last condition on the line marked # ***
Usage:
> kendallTauDistance(c(1,2,4,3),c(2,3,1,4))
[1] 3
You could use
(choose(length(x),2) - cov(x,y,method='kendall')/2)/2
if you know that both of the input lists x and y do not contain duplicates.
Hmmm. Somebody is interested in exactly same thing which I have been working on.
Below is my code in python.
from collections import OrderedDict
def invert(u):
identity = sorted(u)
ui = []
for x in identity:
index = u.index(x)
ui.append(identity[index])
print "Given U is:\n",u
print "Inverse of U is:\n",ui
return identity,ui
def r_vector(x,y,id):
# from collections import OrderedDict
id_x_Map = OrderedDict(zip(id,x))
id_y_Map = OrderedDict(zip(id,y))
r = []
for x_index,x_value in id_x_Map.items():
for y_index,y_value in id_y_Map.items():
if (x_value == y_index):
r.append(y_value)
print r
return r
def xr_vector(x):
# from collections import OrderedDict
values_checked = []
unorderd_xr = []
ordered_xr = []
for value in x:
values_to_right = []
for n in x[x.index(value)+1:]:
values_to_right.append(n)
result = [i for i in values_to_right if i < value]
if(len(result)!=0):
values_checked.append(value)
unorderd_xr.append(len(result))
value_ltValuePair = OrderedDict(zip(values_checked,unorderd_xr))
for key in sorted(value_ltValuePair):
# print key,value_ltValuePair[key]
ordered_xr.append(value_ltValuePair[key])
print "Xr= ",ordered_xr
print "Kendal Tau distance = ",sum(ordered_xr)
if __name__ == '__main__':
print "***********************************************************"
print "Enter the first string (U):"
u = raw_input().split()
print "Enter the second string (V):"
v = raw_input().split()
print "***********************************************************"
print "Step 1: Find U Inverse"
identity,uinverse = invert(u)
print "***********************************************************"
print "Step 2: Find R = V.UInverse"
r = r_vector(v,uinverse,identity)
print "***********************************************************"
print "Step 3: Finding XR and Kenday_Tau"
xr_vector(r)
About the approach/ algorithm to find Kendall Tau distance this way, I would either leave it to you, or point towards the research paper Optimal Permutation Codes and the Kendall’s τ-Metric
You can implement (Approach) the same in R.