Motaz_Husain's blog

How can I solve the "lucky?" Problem ?

By Motaz_Husain, 2 years ago, In English

I Had A Problem solving the "lucky?" problem. And I Need just a little bit of help to solve it.

--> this is the link to the problem : https://mirror.codeforces.com/contest/1676/problem/A

--> and this is my source code :

include

include

using namespace std;

int main(){ string num; int testCases,sum1,sum2;

cin>>testCases;

for(int k=0;k<=testCases-1;k++){

    cin>>num;

    for(int i=0;i<=2;i++){
        sum1=sum1+num[i];
    }//end for i - first 3-digits -

    for(int j=3;j<=5;j++){
        sum2=sum2+num[j];
    }//end for j - Second 3-digits -

if(sum1==sum2){
    cout<<"YES\n";
}//end if - sum1==sum2 -
else {
    cout<<"NO\n";
}//end else

}//end for k-test cases-

return 0;

}//end main

So If Anybody can help I Would be super appreciative.

  • Vote: I like it
  • +4
  • Vote: I do not like it

Comments Comments (9)
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2 years ago, # |
  Vote: I like it +9 Vote: I do not like it

Initialise sum1=sum2=0

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

As the constraints are very limited you can just compare the sum for first 3 to the last three as s[0]+ .. +s[2] to s[3]+ .. + s[5], and have your answer. :)

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Just want to point out: Of course you are right, also with your approach you don't need to initialise sum1 and sum2 and probably I would've done the same in a contest. But I really like, that Motaz_Husain used a for loop! This is very good form in my opinion and especially useful for someone who is starting into CP.

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2 years ago, # |
← Rev. 2   Vote: I like it +12 Vote: I do not like it

To help you helping yourself: find out about the debugging tools of your programming environment or use some temporary debug outputs. By inspecting sum1 and sum2 you would've found, that there are invalid values stored in them.

Edit: Also please use the formating options CF provides.

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2 years ago, # |
← Rev. 2   Vote: I like it -8 Vote: I do not like it

Keep it simple.

string s;
cin >> s;
cout << (s[0]+s[1]+s[2]==s[3]+s[4]+s[5] ? "YES" : "NO") << '\n';
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2 years ago, # |
← Rev. 2   Vote: I like it +8 Vote: I do not like it

i think if u use % and / will be better for u

how to do it

try it in paper it's easy idea

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»
2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

while (t--) {

str s; cin >> s;

if (((int)s[0] + (int)s[1] + (int)s[2]) == ((int)s[3] + (int)s[4] + (int)s[5]))

cout << "YES" << endl;

else

cout << "NO" << endl;

}

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