[BZOJ3823] [East! Simulation competition _ Round5T1: sentiment message push formula + linear screening inverse element (simplified comparison of the public method), bzoj3823_round5t1

[BZOJ3823] [East! Simulation competition _ Round5T1: sentiment message push formula + linear screening inverse element (simplified comparison of the public method), bzoj3823_round5t1

Question 1:

We define a point as a zero-dimensional element, a line as a one-dimensional element, and a plane as a two-dimensional element ......

Since a low-dimensional supercube is translated on the corresponding new axis to obtain a high-dimensional supercube, for example, a two-dimensional supercube is a plane, and then expands along the new Z axis, then a low-dimensional element will increase the one-dimensional element into a high-dimensional element, such as the point to the line, line to the surface, and surface to the body ......

In this way, there is a push type:

High-dimensional elements are derived from the first dimension, and are replicated and retained by the same dimension elements,

That is to say, after a cube is dimension-increasing, the number of I-dimensional elements of the high-dimensional supercube will be the number of I-dimensional elements of the original supercube * 2 + (I-1.

Then, a formula for the combined number can be obtained through the derivation/collision calculation of the ghost animals,

Finally, add a linear screening inverse element, and pay attention to the memory.

Question 2:

Starting from the relationship between the degrees of elements, there is a rule:

The number of X-1 dimension elements corresponding to an x Dimension Element is fixed, for example, a line is always two points, a surface is always four points ...... The I Dimension Element corresponds to 2i (I-1) dimension elements. The number of x + 1 dimension elements corresponding to the x dimension increases with the increase of dimensions. For example, the two-dimensional supercube has a point and two lines, and the three-dimensional cube has 1.3 lines, that is:

An x-Dimension Element corresponds to (y-x) (x + 1) dimension elements in y-dimension.

Then, based on the doubling of the number of points, we can get the number of zero-dimensional elements (points) of the cube, and then export the number of higher-dimensional elements according to the formula, in this way, the O (n) solution can be obtained. However, due to various modulo operations, a linear screening inverse element is required ~~

This is my method, but unfortunately ...... The fast power long is written as int !!!!! Weak.

Code:

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define N 10001000using namespace std;int n,p;long long A[2];long long inv[N];long long power(long long x,int k){long long ans=1;while(k){if(k&1)ans*=x,ans%=p;x*=x,x%=p;k>>=1;}return ans;}void getinv(int x){inv[1]=1;for(int i=2;i<=x;i++)inv[i]=(long long)(p-p/i)*inv[p%i]%p;}int now,last;int main(){int i,j,k;scanf("%d%d",&n,&p);now=0,last=1;A[0]=power(2,n);getinv(n);long long ans=A[0];for(i=1;i<=n;i++){now^=1,last^=1;A[now]=A[last]*(n-i+1)%p*inv[i]%p*inv[2]%p;ans^=A[now];}cout<<ans<<endl;fclose(stdin);fclose(stdout);return 0;}