Comments by jatinp510 (Top 36 by date)

sir, i have tried on my level.
if someone gave me suggestion.
i just want to know how and why creating char array will help.

i just need explaination.
u r expericed peoples and can share your knowlenge with this noob.

sooo... is it crime to ask why for any query

somebody suggest me to
create char array.
how is it done.
and why its neends to be done.

i guess double can't store value of 10^100.

doubles don't have that much range.

not working

sir....
loves ur way to teach.
Always do research...
thnx...

i don't lnow how to do this.
plzz give me soln.

MY CODE IS
#include <bits/stdc++.h>
using namespace std;


int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long t,n,ans;
cin>>t;
while(t--)
{
cin>>n;

ans = ((2%1000000007) * (pow(3,n)%1000000007))%1000000007;


cout<<ans-1<<endl;



}
return 0;

}

||=== Build file: "no target" in "no project" (compiler: unknown) ===|
/home/jatin/programs/tri.cpp||In function ‘int main()’:|
/home/jatin/programs/tri.cpp|16|error: invalid operands of types ‘__gnu_cxx::__promote_2<int, long long int, double, double>::__type {aka double}’ and ‘int’ to binary ‘operator%’|
||=== Build failed: 1 error(s), 0 warning(s) (0 minute(s), 0 second(s)) ===|

sir.. waiting for your reply.
plzz refer to the above code once.

plzzz... refer to the ques in hackerearth.com ...
the user is restricted to use only 5 characters input if n=5.
he can't use 6 characters.
and if my code is wrong for arr[n];
then,
why it is working for 1 test case of the ques.

here i am changing the size of array in every test case

#include<bits/stdc++.h>
using namespace std;

int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);

int t,n,i;
cin>>t;
while(t--){
cin>>n;
int a[n],b[n];
for(i=0;i<n;i++){
cin>>a[i];
}
for(i=0;i<n;i++){
cin>>b[i];
}
sort(a,a+n);
sort(b,b+n);

if(a[n-1]==b[n-1])
cout<<"Tie\n";
else if(a[n-1]>b[n-1])
cout<<"Bob\n";
else
cout<<"Alice\n";
}
return 0;
}

https://www.hackerearth.com/practice/basic-programming/implementation/basics-of-implementation/practice-problems/algorithm/warcakewalk/

#include <stdio.h>

int main()
{
int t,n,i;
scanf("%d",&t);

while(t--){
scanf("%d",&n);
char arr[100000];// what if i write char arr[n]
int count=0;

scanf("%s",&arr);
for(i=0;i<strlen(arr);i++){
if(arr[i]=='0')
count++;
}

printf("%d\n",count);
}
return 0;
}

what is the problem with changing the size of array in every test case.
and why to use dynamic memory if we have lots of memory.

thank u..

got it... Thank you..

can u give me soln of this ques with int type.
And thanku :)

which data type i have to use to hold value for 10^100.

Okay. :)

ITS STILL NOT WORKING

give me some hints

sry for that...

#include <stdio.h>


int main()
{
long n;
scanf("%d",&n);
while(n--){
double g;
scanf("%d",&g);
if(((int)g)%2==0)
printf("YES\n");
else
printf("NO\n");

}
}
Constraints
1<=n<=100000
1<=g<=10^100

here n is the no. of test case
and g is the no. to check whether its even or not.
sirr its not working.

can u plz.. send me your code

thnk u... once again :)

sir i m a begineer...
plzz.. give me some precious suggestions.. :)

//in this code i have checked for whether this is palindrome and if its palindrome it will check whwther its odd or even
//suggest me a better code to improve my skills

#include <stdio.h>
#include <string.h>

int main()
{
int n,p,i;
scanf("%d",&n);
while(n--){
int flag=0,type=0;
int j=0;
char s[100000];
scanf("%s",&s);
j=strlen(s);
if(j==1){
printf("YES ODD\n");
}
else{

if(j%2==0){
type=1;
}

for(int i=0;i<j/2;i++){
if(s[i]==s[j-i-1])
flag++;
p=i;}
if(flag==(p+1)&&type==1)
printf("YES EVEN\n");
else if(flag==(p+1)&&type==0)
printf("YES ODD\n");
else
printf("NO\n");

}}
}

//in this code i have checked for whether this is palindrome and if its palindrome it will check whwther its odd or even
//suggest me a better code to improve my skills

#include <stdio.h>
#include <string.h>

int main()
{
int n,p,i;
scanf("%d",&n);
while(n--){
int flag=0,type=0;
int j=0;
char s[100000];
scanf("%s",&s);
j=strlen(s);
if(j==1){
printf("YES ODD\n");
}
else{

if(j%2==0){
type=1;
}

for(int i=0;i<j/2;i++){
if(s[i]==s[j-i-1])
flag++;
p=i;}
if(flag==(p+1)&&type==1)
printf("YES EVEN\n");
else if(flag==(p+1)&&type==0)
printf("YES ODD\n");
else
printf("NO\n");

}}
}

thnk u.. sir

thnk u soo much