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C++, Competetive program It's hard for me to crack this problem Help me .
write a c++ program using iostream that asks a user to enter 10 integers in an array? It then asks for a number to search and prints all the locations where the number is present? Ans-> As i try to made the program but i am stuck my program taking input properly and when i ask user to search a specific number its also working fine but what if we have multiple number in same array how can we find the specific number in an array and print their index number respectively
7 Answers
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"What if we have multiple number in same array"
Just remove the `break` statement in the for...loop you used to find indices of matching elements.
The task didn't specificly tell you to print "Sorry you have enter wrong number" did it? cause unnecessary outputs may fail your attempt in such challenges. Outputs must *strictly* follows the instruction, no more and no less.
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No bro, no dumb, just forgot some things LOL
Next time post your code bit link ok bro đ
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Bro you can create new code in Playground, paste your code copied from your laptop IDE. And then you can save it, and share the link, just follow this easy steps đ
https://www.sololearn.com/post/75089/?ref=app
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Ipang how dumb I am lol thanks again bro đ
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Ipang I don't know how to send code by link , either I write the whole code or I copy paste from laptop ide
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Ipang got it .
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#include<iostream>
using namespace std;
int main(){
int array[10],i;
//Asking from user to enter number
cout<<"Enter 10 number to submit "<<endl;
for( i=0;i<10;i++)
{
cin>>array[i];
}
//printing the number user have submit
cout<<"The number you have entered is"<<endl;
for( i=0;i<10;i++){
cout<<array[i]<<endl;
}
//Asking for any random and return their index number
int num;
cout<<"Enter the number which you want to know their location"<<endl;
cin>>num;
for( i = 0; i < 10; i++){
if(array[i] == num){
cout << "Element found at index " << i;
break;
}
}
for(i=0;i<10;i++){
if(array[i] != num){
cout<<"Sorry you have enter wrong number"<<endl;
break;
}
}
return 0;
}