# Logic for solving Guess It Right problem on CodeChef?

The problem was asked on febuary 2019 long challenge.I was able to solve the problem partially using two loops but i got TLE for last two test cases.
Can someone share the right approach.Thanks

Basically - two approach are possible -
1.Chef removes the boxes after every move (if it is not last move ).
2.or greedy approach.

I am new to Competitive programming.Also i tried this problem for long time.

If CHEF has more than 1 move then he will first remove boxes then he will choose one box. This leads to a GP
\frac{1}{n} + (1- \frac{1}{n})*\frac{1}{n}+(1- \frac{1}{n})^2*\frac{1}{n}+....
If m is odd then GP would have \frac {m}{2} terms. If m is even then still GP would have same \frac {m}{2} terms and also a last term would be {(1- \frac{1}{n})}^{m/2}*\frac{1}{n+k}. So summation of this GP is the answer.
If m is ODD then answer is
\frac{1}{n} + (1- \frac{1}{n})*\frac{1}{n}+(1- \frac{1}{n})^2*\frac{1}{n}+...+{(1- \frac{1}{n})}^{m/2}*\frac{1}{n}
Else (m is EVEN) answer is
\frac{1}{n} + (1- \frac{1}{n})*\frac{1}{n}+(1- \frac{1}{n})^2*\frac{1}{n}+...+{(1- \frac{1}{n})}^{m/2-1}*\frac{1}{n} + {(1- \frac{1}{n})}^{m/2}*\frac{1}{n+k}

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Hi @jbhv12 ! I modified your solution and submitted. It got an AC. See here.
Mistakes i found in your code:-

1. Instead of power you used inbuilt pow function many times. That would obviously result in overflow and give wrong value. I replaced every pow with your power().

2. I replaced your following line ll p = (pow(n,y+1)) - (n*pow(n-1,y)) + (pow(n-1,y));
with

p -= (n*power(n-1,y, mod))%mod;
p += mod;
p %= mod;
p += (power(n-1,y, mod))%mod;
p %= mod;

Because whenever you subtract something then you may get negative value so taking modulo will also result in negative answer. So always do ans = (ans+mod)%mod.

ll p1 = a*q + p*b;
with ll p1 = (a*q)%mod + (p*b)%mod;

4) and put %mod on almost every step.

Hooray! It resulted in AC.

General Tips for such questions:

1) Always use modulo power exponentiation whenever answer is asked to compute modulo M.

2) Whenever ans becomes negative(either by subtracting or by multiplying with negative number) use ans = (ans%MOD + MOD) % MOD as suggested by @l_returns

3) Use %MOD after each arithmetic operation.`
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@rajput1999 could you please explain from scratch (atleast name the recommended topics)

@vichitr I figured out exact same logic. Here’s my code getting WA: https://www.codechef.com/viewsolution/22948884

I’ve been pulling my hair. help me?

@vichitr Thanks for answer , it was really helpful .How do i prove that removing the boxes after each move is best option. obviously its feels that it should be but how do i prove it.Thanks

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if you don’t understand anything feel free to ask

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Thanks @rajput1999

correction :
2) Whenever ans becomes negative(either by subtracting or by multiplying with negative number) use ans = (ans%Mod + MOD) % MOD