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For above input your code produces though the output should be Your arrays aren't global, you've declared them inside main().

Try this input You might get run-time error. Declare the char arrays as static or global.

Yes, since 0 has already appeared we terminate the chain, not including the last 0 since it's already there. So the cycle contains only one number.

So it's the streets that must ensure that at least one end has a center. Yes, my previous assumption would have missed streets (4 6) and some more. Thank you Hisoka.

Probably I haven't understood the problem clearly. Aren't we supposed to find the minimum number of information centers needed to form so that all the junctions have a center either at it or adjacent to it? If so, for the graph posted by arsalan_mousavian 6 centers should be enough, I think. Don't 1...

You might like to try the inputs given in earlier posts.

Programmer wrote:Hello.
Can someone who get AC tell me answer of this test.
1
1 10000000
Thanks.

Case 1: 977787

You may try this inputs

There might be something wrong with your input. Anyway I get the same output as yours.

Input Output

Thank you so much Krzysztof Duleba. You're damn expert in C++.

Thank you Krzysztof Duleba. I've rewritten sq_distance() as But if I try to pass the arguments by const references to cross_product() as well then I come with new compile errors which seems to be related to the operator-().

Thanks. :) Changing bool comp(point &a,point &b) bool eqal(point &a,point &b) inline int sq_distance(point &p1,point &p2) inline int cross_product(point &p0,point &p1,point &p2) to bool comp(const point &a,const point &b) bool eqal(const point &a,const...

My following code is getting compile error. Can anyone help me in this reagard? #include <cstdio> #include <vector> #include <algorithm> using namespace std; #define SIZE 100000 struct point{ int x,y; point(){} point(int _x,int _y):x(_x),y(_y){} point operator-(point &p){point tmp(x-p.x,y-p.y);r...

jainal uddin wrote:hello try to make another algorithm to find out humble number.

Are you answering yourself!?