cf#345 (DIV1)

The puzzle after ruin

First spit out your own code ability & hand Speed

A 06:52 hand in.

B 29:34.

C 52:49.

Everything's fine.

Then around 01:15 found that C looks like 100W will explode stack, scared me to change to VC + + opened the stack re-handed over again

So hurriedly go to room sweep a wave, and then found a group with DFS, feel very happy, start hack, and then success-2 ... Finally, the system test is over ...

I have to hurry to see how big the CF stack space is.

UPD:

I later CF do the problem again no matter the stack space ... System Stack Default 256M ...

And then 01:46 when found a a place did not open long long incredibly be people hack ... That's a conscience hacker.

Lying trough my punishment when ah ... hurriedly re-handed a copy of a 01:57 ...

After the final game found B false negative a situation ... So far officially GG

On how Konjac Konjac is being abused by CF

Anyway, I am still very happy, as long as the blue will be able to play a happy Div2 (Div2 is the normal CF that can ak,hack relaxed pleasure) Unfortunately there is no blue

By the way, I glanced at Div2 's question.

The following are the ===================================== of the discredited bacteria.

Div2 A.joysticks

The main idea: two good friends to play games, they have two joystick a charger, began the first charge of the a1%, the second charge of the power of the a2%, every minute to play the game if a pole inserted electricity will be +1% of the electricity, not plug will-2% of the electricity. When a pole is out of power (0%), they can't play the game. A pole can charge more than 100% of the power. (Don't ask me why) ask them how long they can play the game.

The puzzle: Note that the total amount of electricity will not increase, then we charge less per minute simulation can.

//div2a#include <iostream>#include<stdio.h>#include<stdlib.h>#include<algorithm>#include<string.h>#include<vector>#include<limits>#include<Set>#include<map>using namespacestd;intMain () {inta1,a2,t=0; CIN>>a1>>A2;  while(a1&&a2&& (a1>=2|| a2>=2))    {        if(A1&GT;A2) ++A2, a1-=2; Else++A1, a2-=2; ++T; } cout<<t;}

In construction

cf#345 (DIV1)