_PsychoKiller's blog

By _PsychoKiller, history, 3 years ago, In English,

OK, so i didn't get what the tutorial of that really meant, and that tutorial looked long..-_-

Fortunately, I found a better solution. And my solution is mainly like 2 lines!!! :p I think it's better if you see the code rather than me trying to explain it.....

include<bits/stdc++.h>

using namespace std;

int main(){ int n, i; cin>>n; for(i=0;i<n;i++)cout<<i+n*10<<" "; return 0; }

For any question or suggestion, feel free!!!

 
 
 
 
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3 years ago, # |
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Auto comment: topic has been updated by 1212 (previous revision, new revision, compare).

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3 years ago, # |
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To be very simple You just have to print N Prime Numbers less than 10^7 .

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3 years ago, # |
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Do you seriously concatenate every line and believe that you have written only one line of code? :-/

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4 months ago, # |
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A very simple solution would be printing n consecutive numbers starting from 10^5 . The next number divible will be 2*10^5 which is 10^5+1 characters from 10^5 thus stisfying the condition.