"Simulation" Rabbit Farm

Free farming CH Round #54-streaming #5 (Noip simulation Day1)Describe

(If you want to better understand the subject, please read NOI2011 first try "Rabbit Farm" a problem)
When she was worried about how to make more money in recent years, she heard the children next door talking about the problem of child-free reproduction. (Note: Child-free is a simple single-celled organism)
The problem is this: time 0 has 2 newborn children. Every moment, each child will be divided into 2 free children. Q Time n How many free children?
Smart you may have found that the moment N of the number of n+1 is exactly the power of the first 2. Meng eggs do not understand what is called power, but she also found the law: the moment n+1 the number of free children equal to the time n the number of free children twice times. The number of free children in the first few moments (starting from 0) is:
2 4 8 16 32 64 128 256 512 ...
Meng eggs found more to the back of the increase in the number of children, expect to raise the son must be able to make a lot of money, so the egg at the moment 0 bought 2 free children began to cultivate.
Every day, the eggs are to give the children to provide nutrition. The medium of the child-free is very special, each K-free child occupies a medium, the last remaining less k occupies only one medium. Because the child is particularly afraid of loneliness, if a medium only 1 free of children, the child will soon die.
However, the number of free children at each moment can still be calculated. For example, when k=7, the number of free children from the first few moments (starting at 0) is:
2 4 7 14 28 56 112 224 448 ...
Given n, you can help the egg to calculate the moment n how many children do she have? Because the answer can be very large, you just need to tell the egg-n-free number to the remainder of P.

Input format

The input has only one row and contains three integers n k p.

Output format

The output has only one row, which is an integer that represents the remainder of the time N's free-of-number pairs p.

Sample input

6 7 10086

Sample output

112

Data scope and conventions

For 30% of data, n≤1,000,000.
Another 30% of the data, K is the positive integer multiples of p.
for 100% of data, n≤1,000,000,000,2≤k≤1,000,000,1≤p≤1,000,000.

#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <string > #include <cmath> #include <algorithm> #include <queue> #include <vector> #include <set >using namespace std; #define INF 1<<30#define LL long longll n,k,p; ll Pow (ll x) {      ll ans=1,y=2;      while (x)      {          if (x&1) ans=ans*y%p;          Y= (y*y)%p;          x=x>>1;      }      return ans%p;} int main () {      scanf ("%i64d%i64d%i64d", &n,&k,&p);      LL ans=2,tem=2;      if (k%2==0)      {            printf ("%i64d\n", 2*pow (n));            return 0;      }      for (int i=1;i<=n;i++)      {         tem= (tem*2)%k;         if (tem%k==1)         {            printf ("%i64d\n", (ans*2-1) *pow (n-i)%p);            return 0;         }         Ans= (ans*2)%p;      }      printf ("%i64d\n", ans);      return 0;}

　　

"Simulation" Rabbit Farm