Try this test:
100000 99999 0
1 2 1000000
2 3 1000000
...
99999 100000 1000000

Answer: 10049708502000000

Good Luck!

Thanks!

HINT for solvers: In the test above, you can get an overflow. Change min(⌈(1 + T/100)* ti⌉, 100500*ti) to
long long my_ceil(long long tnow, long long f1, long long t)
{
    if(h1+tnow-f1>h2*h1) return h2*t;
    //return min(ceil(t+(tnow-f1)*t/100.0), h2*t);
    if(((tnow-f1)*t) % h1==0) return t+((tnow-f1)*t)/h1;
    else return t+((tnow-f1)*t)/h1 + h3;
}
//h1=100; h2=100500; h3=1;
where tnow=time now, f1=last cleaning time for this rib, s2=start time of next cleaning


Edited by author 26.07.2017 14:10