OJ Board

Thanks for your help, but it seems to me, that answer for case 6: 4 1 2 3 3 1 5 1 1 4 3 4 2 will be 3 1 1 2 1, but not 0 1. I think 3 1 1 2 1 divide both 4 1 2 3 3 1 and 5 1 1 4 3 4 2 with no remainder. Plz explain. PS. I've just changed module of ring 1500 for prime number(53) for correctness of in...

or look at my bad code(sorry for bad style): #include <stdio.h> #include <memory.h> #define FOR(i,n) for(i=0;i<(n);i++) #define FORR(i,a,b) for(i=(a);i<=(b);i++) #define set(x,f) memset(x,f,sizeof(x)) int n,i,j,nt,bf; int a[103],b[103],c[103]; int r[1501]; void mod(int a[103],int (&b)[103]) { in...

Someone who have gotten AC give me plz some test cases

2 trulo17
Why you add to angel pi?
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To solvers: Help us plz.

I always get TLE on this problem.
Using brute forse on each step I try go from one diamante to another or go to exit(when i go to exit i find another unused diamonde and use recursion again).
How can i prune it?
Give me some hints.Plz.

Hello! Can anyone give me several tests for 10492 with answers? plz.

What kind of C++ compiler(version) is using by Online-Judge?
And where can i download it?
Plz help