Yeah, really nice idea. I will implement it next time. It's partially shown on the second graph (accepted/rejected/hacked), but you are right, it will be better shown with numbers too.

In this round it was: A: 64.36 % (nice!) B: 5.88 % C, D, E with 0%.

Does anyone have an idea for this? I thought about locking this problem, because I thought some people might miss the subtraction with mods, but I couldn't find a good way to exploit it. It turned out no one in my room failed because of it, but still...

Nevermind, I got it. First assume the string is of some convenient form, like bbb...b+acca+bbb...b (acca within a lot of b's), and the string to search for is "a". For this just try things (I tried a single "a" within before, and also "aca" within). Let there be x b's at the beginning and y b's at the end. Then, the answer is the sum of 3(x + 1) + 3(y + 1) for the one element sequences with 1 "a" and no "c"s, (x + 1)(y + 1) for the one element sequences with both "a"s, and 6(x + 1)(y + 1) for the two element sequences, giving 3(x + 1) + 3(y + 1) + 7(x + 1)(y + 1). W|A (why doesn't my link work? see below for link) gives integer solutions, for example x=3366,y=84845.

This works:

for (int i=0;i<3366;i++) printf("b");
printf("acca");
for (int i=0;i<84845;i++) printf("b");
printf("\na\n");

Try hacking this solution: 9106264.

Go to wolframalpha.com and search 3*(x+1)+3*(y+1)+7*(x+1)*(y+1)=2000000013, 0<x<100000, 0<y<100000, x+y<100000.