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Wipro, Tcs, MindTree and Accenture Solution Group:
Accenture Exam (Questions)Answer
This is the format 👉👉
( Questions ) Answer
1, Speak
2, (higher studies)On
3, ( Incensed)Affected
4, (games) play
5, (Loved) Hated
6, (Sports) sisiter still at
7, (leaves of) Both A and B
8, (United Kingdom) The
9, (water) Drinking
10, (myson) of
11, (interim) Temporary
12, (Come Late) sent word
13 (Denigrate) Extol
14, (indian cardamom ) better in quality
15, ( unemployment) No correction
16, (diamond corundum) Ruby
17, (belong to that group) 426
18, (Comes inside the circle) 6
19,(3, 289) 3414
20, (Brick, cement sand) wall
21( 22,6) 31
22( Train is called) Tractor
23, (assurance is not genuine) only assumption 2 impl...
24, (Chapati) Karnataka
25,(south:north: west) North East
26,( application of appli) If Both 1 and 2 follo
27,
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+2Accenture Exam
English ability
1.kids when
B competition
2.usually social
B Social workers tend
3.in a row of frds
B 12
4.Paul is
D 38
5.Ranbir
D non of these
6.Five persons
A banglow
7.Alexa checks
B Tuesday
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#include<iostream.h>
#include<ctype.h>
void main()
{
int count=0;
char string[10];
cin>>string;
for(int i=0;string[i]!='\0';i++)
{ if(isupper(string[i])) count++; }
cout<<"The Number of Upppercase letters is"<<string;
}
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CAPGEMINI, ACCENTURE & COGNIZANT Exam Answer
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l=list(map(int,input().split(',')))
m=[]
for i in range(5):
k=[]
for j in range(5):
k.append('O')
m.append(k)
for i in l:
a1=(i-1)//5
a2=(i-1)%5
m[a1][a2]='X'
num=0
for i in m:
x=0
o=0
for j in i:
if j=='X':
x=x+1
else:
o=o+1
if x==5:
num=1
if o==5:
num=2
for i in range(5):
x=0
o=0
for j in range(5):
if m[j][i]=='X':
x=x=1
else:
o=o+1
if x==5:
num=1
if o==5:
num=2
o1=0
x1=0
o=0
x=0
for i in range(5):
for j in range(5):
if i==j:
if m[i][j]=='X':
x1=x1+1
else:
o1=o1+1
if i==5-j-1:
if m[i][j]=='X':
x=x+1
else:
o=o+1
if o1==5 or o==5:
num=2
if x1==5 or x==5:
num=1
print(num)
for i in range(5):
for j in range(5):
print(m[i][j],end='')
if j!=4:
print(',',end='')
print()
Uni- Directed Tree Node Code..
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ans = 0; XOR = 0;
prefix = [0] * n;
for i in range(n) :
XOR = XOR ^ arr[i];
prefix[i] = XOR;
oddGroup = dict.fromkeys(prefix, 0)
evenGroup = dict.fromkeys(prefix, 0)
oddGroup[0] = 1;
for i in range(n) :
if (i & 1) :
ans += oddGroup[prefix[i]];
oddGroup[prefix[i]] += 1;
else :
ans += evenGroup[prefix[i]];
evenGroup[prefix[i]] += 1;
return ans;
Sub Array count code
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All EXAM SOLUTION GROUP:
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