OJ Board

I used the following "fix" function: double fix(double v) { if (v>0) return v; if (v>-0.0005) return 0.0; return v; } and printed the complex numbers like this: for(i=0;i<n;i++) { printf("%0.3lf",fix(cplx .re)); if (fix(cplx .im)>=0) printf("+"); printf("%0.3lfi\n&...

Problems with multiple input where each test case end with end-of-file (that is, a blank line in the multiple input format!) is really nasty. On such problem, I always read the whole input using gets(s) and then sscanf(s,...) and so on. To detect the end of a testcase, check (feof(stdin) || s[0]) A ...

You must declare it like this:

const char check[]= "`1234567890-=QWERTYUIOP[]\\ASDFGHJKL;'ZXCVBNM,./ nn";

Remember that \ has a special meaning in strings!

You don't seem to handle the new line character! It should also be echoed.

*sigh* I totally forgot that the string compare should be reversed since the prefix is appended backwards.

I keep getting WA, I'm not sure why the following algorithm doesn't work: 1) If strings equal, print "No solution." 2) If one string is empty: 2.1) If the other string is not a palindrome, print blank line 2.2) Otherwise either add 'a' or 'b' (just check which one makes the other string no...

The answer two your two cases are 2 and 1. Treat numbers just like word. Ignore all chars which are not digits or letters.

I believe that the writing CCIC is not really valid - or not a "correct" writing. Either such cases are omitted in this problem, because my program (which was AC) didn't consider them valid, or they are really incorrect. But I guess there are different opinions in this matter.

I guess the easiest way to solve it is to loop through all numbers 1-3999 and generate what that number is in Roman numerals (a much easier task), and then do a table lookup. A linear search even worked, but that was slooow

That input is draw, that is correct. I start by looping through all 2*32*32*32 states (at least half the squares are filled with #) and mark any state with M at a phone as a win and M at an A as a loss (in case of both of these at the same square, it doesn't matter since this position isn't reachabl...

Your program doesn't work very well on large inputs, for instance

1000000 34238

In this problem you can handle big inputs in a special way; if there are enough stones it's always possible to reach a certain stone. Why this is so I don't want to prove

That looks correct although I prefer to use the evaluation symbols +1 (Good Guy win), 0 (draw) and -1 (Bad Guys win) all the time and not +1 if the guy to move will win whether or not he is Good or Bad (I think that is more confusing). The graph does contain loops which is why you must search backwa...

It can probably be quite high. Use dynamic allocation, I do something like this:

char *sq;

while (scanf("%d",&m)==1) {
sq=malloc(m*m+1);
for(i=0;i<m;i++)
scanf("%s",sq+i*m);
...
}

I've done _many_ submissions on this one now, always WA. I do a BFS on the polygon (together with which edges have been dipped already), "shrinking" one edge being a move. I am certain the polygon clipping routine works, it's been tested a lot. So are there any other things I'm missing? So...

Nice long message, eh?

The output should be

2
1
3
1
1
3
2
7
1
1
4
5
5
7
8
5
1