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| What could be the purpose of taking N integers as string. |
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Doctor Cop
Member #16,833
April 2018
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1#include<iostream>
2using namespace std;
3int main(){
4 int T;
5 cin>>T;
6 for(int a=1;a<=T;a++){
7 int result =0;
8 string s;
9 cin>>s;
10 int n=s.size();
11 for(int i=0;i<n;i++){
12 result =result +s[i];
13 }
14 if(result%3==0){
15 cout<<"Yes"<<endl;
16 }
17 if(result%3!=0){
18 cout<<"No"<<endl;
19 }
20 }
21 return 0;
22}
This is the code I found in the submissions after trying four time in a row with my simple logic. This is something I am unable to wrap my head around, that much I know that my logic didn't work because the inputs were very large but how is this code not having any problems with size limit? This is the problem:- 1You are given a number N you need to check if it is divisible 3.
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3Input:
4
5The first line of the input contains an integer T, denoting number of test cases. Each of the next T lines contains an integer N.
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7Output:
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9For each of the test cases print "Yes" if N is divisible by 3 else "No".
10
11Constraints:
12
131<=T<=10^3
14
151<=N<=10^1000
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GullRaDriel
Member #3,861
September 2003
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This will overflow like any other. result =result +s[i];
Edit: "Code is like shit - it only smells if it is not yours" |
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Peter Hull
Member #1,136
March 2001
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This is taking account of the divisibility rule - you just need to add the digits and check that's divisible by 3. Since the number has 100 digits or less there's no risk its sum of digits will overflow. Really result =result +s[i] should be result = result + s[i] - '0' but I think it won't matter because the ASCII code for '0' is itself divisible by 3 (48 is it?)
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GullRaDriel
Member #3,861
September 2003
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In the given example there can be an overflow because T is not capped like Peter said, and like the constraint was asking. "Code is like shit - it only smells if it is not yours" |
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Audric
Member #907
January 2001
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These problems give the 'constraints from the outside', in order to plan for worst case scenario. |
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