How the code passes 19 th test case 1 1 1 1 1 1

in my pc it gets RTE!!! its very mysterious.

I have looked through assembly code generated by the compiler and I think I've found the reason. If the first loop in the code is never executed(e.m k == n), it jumps over all the block of code until the loop from 1 to k. So the division does not happen at all. It works only with -O2(or similar) optimization options. Without it this solution gets RE on test 1 1 1 1 1 1.

Amazing!!!!! Why Above code didn't get RTE ?? Testcase 11 was sufficient for getting RTE !!! Almost same solution Get RTE on test 11 due to division by 0. http://codeforces.com/contest/369/submission/5311155

I don't know, it fails on my compiler too.

yo.

an usual for cycle

for(init; condition; increment)
{
    body;
}

an usual assembly for this is:

init;
if(!condition)
    jmp end;
label again:
body;
increment;
if(condition)
    jmp again:
label end:

so basically the conditional check is implemented 2 times (depending on compiler flags). (the compiler cound have used a jmp check: after the init, to jump right to the end, to the conditional, and jump back if it needs to start the cycle.) if there is 2 copies of the conditionals, the first means do we need to start the cycle. in this case initialization of variable tp might be delayed until this point, because tp is only used inside this cycle.

later it will be overwritten, previous value does not matter. compilers are too smart i guess :)

when it comes to here:for(int i=k+1; i<=n; i++) because k+1>n so the pc will ignore the body and jump to next. so it won't get RE.... BTW I find that if n and k are double, it won't get RE either.