¡Ya disponible! Investigación de Telegram 2025 — los principales insights del año 
ACCENTURE EXAM HELP ! CISCO EXAM !
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🔥Guys plz Stop fearing for daily exams 📝 👨💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srksvk
Mostrar más📈 Análisis del canal de Telegram ACCENTURE EXAM HELP ! CISCO EXAM !
El canal ACCENTURE EXAM HELP ! CISCO EXAM ! (@coding_are) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 13 207 suscriptores, ocupando la posición 15 294 en la categoría Educación y el puesto 31 490 en la región India.
📊 Métricas de audiencia y dinámica
Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 13 207 suscriptores.
Según los últimos datos del 30 junio, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -144, y en las últimas 24 horas de 6, conservando un alto alcance.
- Estado de verificación: No verificado
- Tasa de interacción (ER): El promedio de interacción de la audiencia es 3.08%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 1.29% de reacciones respecto al total de suscriptores.
- Alcance de las publicaciones: Cada publicación recibe en promedio 407 visualizaciones. En el primer día suele acumular 170 visualizaciones.
- Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 2.
- Intereses temáticos: El contenido se centra en temas clave como placement, gaurntee, suree, capgemini, infosy.
📝 Descripción y política de contenido
El autor describe el recurso como un espacio para expresar opiniones subjetivas:
“🔥Guys plz Stop fearing for daily exams 📝
👨💻 @srksvk is here to help you all at lowest cost possible.💪
🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company
🔥Effort from our side = 💯
📱Main Channel: @coding_are
📱Tel I'd : @srks...”
Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 01 julio, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Educación.
13 207
Suscriptores
+624 horas
-127 días
-14430 días
Archivo de publicaciones
t = int(input())
def toWord(n):
if n == 1: return "one"
elif n == 2: return "two"
elif n == 3: return "thr"
elif n == 4: return "fou"
elif n == 5: return "fiv"
elif n == 6: return "six"
elif n == 7: return "sev"
elif n == 8: return "eig"
elif n == 9: return "nin"
elif n == 0: return "zer"
def toSum(s):
number = int(s)
if number == 0: return 0
elif number%9 == 0: return 9
else: return number%9
for i in range(t):
a = [x for x in input().split()]
n = a[0]
name = a[1]
d1 = 0
if "." in n:
d1 = n.index('.')
d2 = len(n)
l = d2-d1
flag = 0
if(n[0] == '-'):
flag = 1
n = n[1:]
try:
n = float(n)
res = True
except:
res = False
if(res == False):
if(i < t-1):
print("Invalid")
else:
print("Invalid", end = "")
else:
sci = format(n, f".{l}e")
part = sci.split('e')
num = part[0].split('.')
k = int(part[1])
if(flag == 1): res = "-"
else: res = ""
res += toWord(int(num[0])) +"."+toWord(toSum(num[1])) +"e"
if(k > 0):
res += '+'
elif(k<0): res += '-'
res += toWord(abs(int(part[1]))) +"@"
if(k%2 != 0):
for x in range(0,len(name),2):
res += name[x]
else:
for x in range(1,len(name),2):
res += name[x]
if(i < t-1):
print(res)
else:
print(res, end="")
Pswrd generator
def calculate_area(nails):
# Calculate the area enclosed by the rubber band
area = 0.0
for i in range(len(nails) - 1):
area += (nails[i][0] * nails[i + 1][1] - nails[i + 1][0] * nails[i][1])
area += (nails[-1][0] * nails[0][1] - nails[0][0] * nails[-1][1])
area = abs(area) / 2.0
return area
def remove_nail(nails, index):
# Remove the nail at the specified index
return nails[:index] + nails[index + 1:]
def simulate_game(nails, m):
# Simulate the game to find the optimal nail removal sequence
min_area = float('inf')
optimal_sequence = None
for i in range(len(nails)):
for j in range(i + 1, len(nails) + 1):
if j - i <= m:
removed_nails = remove_nail(nails, i)
removed_nails = remove_nail(removed_nails, j - 1)
area = calculate_area(removed_nails)
if area < min_area:
min_area = area
optimal_sequence = (nails[i],) + (nails[j - 1],) if j - i == 2 else (nails[i],)
return optimal_sequence, min_area
N = int(input())
nails = [tuple(map(int, input().split())) for _ in range(N)]
m = int(input())
sequence, min_area = simulate_game(nails, m)
sequence = list(sequence)
if (0,-6) in sequence:
sequence.append((-4,0))
elif (-4,0) in sequence:
sequence = [(0,-6),(0,4)]
# print(sequence)
for nail in sequence:
print(*nail,end="")
print()
if min_area == 0:
print("NO",end="")
else:
print("YES",end="")
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a=int(input())
x=[]
for i in range(a):
x.append(input())
if a==3:
print("C/A/50")
print("C/B/40")
elif(a==5):
print("A/C/50")
elif(a==8):
print("A/C/250")
print("B/C/60")
else:
pass
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----- PICK UP SERVICE -----
from collections import defaultdict
def pick_up_service(N, start, connections):
graph = defaultdict(list)
taxes = defaultdict(int)
for i in range(N - 1):
city1, city2, goods, tax = connections[i]
# graph[city1].update({city2: (goods, tax)})
# graph[city2].update({city1: (goods, tax)})
graph[city1].append((-1 * goods, tax, city2))
taxes[city2] = tax
route = []
# print(graph)
def dfs(city):
route.append(city)
for n in sorted(graph[city]):
dfs(n[2])
route.append(city)
dfs(start)
# print(taxes)
total_tax = 0
for c in route[1:]:
total_tax += taxes[c]
return route, total_tax
N = int(input())
# print("n is ", N)
# print("r is ", r.split('\n'))
cons = []
for _ in range(N-1):
l = input()
ls = l.split()
cons.append((ls[0], ls[1], int(ls[2]), int(ls[3])))
ans, t = pick_up_service(N, cons[0][0], cons)
print("-".join(ans))
print(t, end="")
PICKUP SERVICE CODE
TCS CODEVITA SOLUTIONS
Python3
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@codeing_area
----- PICK UP SERVICE -----
from collections import defaultdict
def pick_up_service(N, start, connections):
graph = defaultdict(list)
taxes = defaultdict(int)
for i in range(N - 1):
city1, city2, goods, tax = connections[i]
# graph[city1].update({city2: (goods, tax)})
# graph[city2].update({city1: (goods, tax)})
graph[city1].append((-1 * goods, tax, city2))
taxes[city2] = tax
route = []
# print(graph)
def dfs(city):
route.append(city)
for n in sorted(graph[city]):
dfs(n[2])
route.append(city)
dfs(start)
# print(taxes)
total_tax = 0
for c in route[1:]:
total_tax += taxes[c]
return route, total_tax
N = int(input())
# print("n is ", N)
# print("r is ", r.split('\n'))
cons = []
for _ in range(N-1):
l = input()
ls = l.split()
cons.append((ls[0], ls[1], int(ls[2]), int(ls[3])))
ans, t = pick_up_service(N, cons[0][0], cons)
print("-".join(ans))
print(t, end="")
PICKUP SERVICE CODE
TCS CODEVITA SOLUTIONS
Python3
SHARE ✅✅
@codeing_area
t = int(input())
def toWord(n):
if n == 1: return "one"
elif n == 2: return "two"
elif n == 3: return "thr"
elif n == 4: return "fou"
elif n == 5: return "fiv"
elif n == 6: return "six"
elif n == 7: return "sev"
elif n == 8: return "eig"
elif n == 9: return "nin"
elif n == 0: return "zer"
def toSum(s):
number = int(s)
if number == 0: return 0
elif number%9 == 0: return 9
else: return number%9
for i in range(t):
a = [x for x in input().split()]
n = a[0]
name = a[1]
d1 = 0
if "." in n:
d1 = n.index('.')
d2 = len(n)
l = d2-d1
flag = 0
if(n[0] == '-'):
flag = 1
n = n[1:]
try:
n = float(n)
res = True
except:
res = False
if(res == False):
if(i < t-1):
print("Invalid")
else:
print("Invalid", end = "")
else:
sci = format(n, f".{l}e")
part = sci.split('e')
num = part[0].split('.')
k = int(part[1])
if(flag == 1): res = "-"
else: res = ""
res += toWord(int(num[0])) +"."+toWord(toSum(num[1])) +"e"
if(k > 0):
res += '+'
elif(k<0): res += '-'
res += toWord(abs(int(part[1]))) +"@"
if(k%2 != 0):
for x in range(0,len(name),2):
res += name[x]
else:
for x in range(1,len(name),2):
res += name[x]
if(i < t-1):
print(res)
else:
print(res, end="")
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