Return-value C++

Question

I'm a new programmer, I was solving a problem at codeforces when I got this error.

http://codeforces.com/contest/342/problem/A

My algorithm works fine but I don't know why at the very end, I have an error message and a return value different from 0. It's shown in the picture below.

https://i.sstatic.net/sz9Jc.png

Thank you in advance !!

Here is my algorithm:

#include <iostream> #include <vector> using namespace std; void print(int a, int b, int c, int n) { for (int i=0;i<n;++i) cout<<a<<" "<<b<<" "<<c<<endl; } int main () { int n; cin>>n; vector<int> V(5); bool cont=true; for (int i=0;i<n;++i) { V[i]=0; } for (int i=0;i<n;++i) { int a; cin>>a; if (a==5 || a==7) cont=false; if (cont==true) { if (a==1) V[0]++; if (a==2) V[1]++; if (a==3) V[2]++; if (a==4) V[3]++; if (a==6) V[4]++; } } if (cont==false) { cout<<-1<<endl; }else { if (V[0]==V[1] && V[1]==V[3] && V[2]==0 && V[4]==0) { print(1,2,4,V[0]); }else if (V[0]==V[1] && V[1]==V[4] && V[2]==0 && V[3]==0) { print(1,2,6,V[0]); }else if (V[0]==V[2] && V[2]==V[4] && V[1]==0 && V[3]==0) { print(1,3,6,V[0]); }else if (V[0]==V[1] && V[1]==(V[4]+V[3]) && V[2]==0) { print(1,2,4,V[3]); print(1,2,6,V[4]); } else if (V[0]==V[4] && V[0]==(V[1]+V[2]) && V[3]==0) { print(1,2,6,V[1]); print(1,3,6,V[2]); } else if (V[0]==(V[1]+V[2]) && V[0]==(V[3]+V[4]) && V[3]==V[1]) { print(1,2,4,V[1]); print(1,3,6,V[2]); } else if (V[0]==(V[1]+V[2]) && V[0]==(V[4]+V[3]) && V[3]==(V[1]/2) && V[2]==(V[4]/2)){ print(1,2,4,V[3]); print(1,2,6,V[1]-V[3]); print(1,3,6,V[2]); } else { cout<<-1<<endl; } } cout<<"fin"<<endl; return 0; }

Does the down-voter care to explain what's wrong with this question? — Hulk
– Hulk, Commented Sep 9, 2013 at 8:56

sara · Accepted Answer · 2013-09-09 08:41:00Z

 for (int i=0;i<n;++i) { V[i]=0; }

You should check that n is 5 or less. Or allocate the vector dynamically. In case of n>5 you are causing heap corruption, which results in the exit code you got.

If you use the value of n for something else and you just wanted to init your vector you can iterate

 for (int i=0;i<5;++i)

Martin Schlott · Accepted Answer · 2013-09-09 08:54:35Z

2

Your vector is to small for the example you show:

vector<int> V(5);

after that line you clean the vector

for (int i=0;i<n;++i) { V[i]=0; }

but in your example you need nine elements. You corrupt the heap with that. Change the first line to

vector<int> V(n);

and it will return the expected 0.

edited Sep 9, 2013 at 8:54

answered Sep 9, 2013 at 8:41

Martin Schlott

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3 Comments

Martin Schlott Over a year ago

same here, downvoting for what?

RobH Over a year ago

We've got a troll lurking.

Zeta Over a year ago

Actually, vector<int> V(max(5,n)) is a little bit more secure, as the code still uses V[4] sometimes.

RobH · Accepted Answer · 2013-09-09 08:37:21Z

The exit code that you see, 3221226356, converted to hex is 0xC0000374. Googling for that indicates that it is the exception code for heap corruption. You've got an error somewhere on cleanup and an exception is thrown, then that exception code has emerged as your process exit code.

Somebody unfamiliar with programming doing some downvoting on this question?
same here, I already flagged myself to get moderation attention.

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