I saw the convolution at a glance, and it intrigued me so much that I realized it was just like this.
Face
Topic: Ask n, 1~m for each number, at least one prime number, and how many sequences are multiples of p, modulus 1e9+7.N < 1e9, m < 2e7, P < 100.
Seeing "there is at least one prime number", which is probably something like repulsion, subtracts the problem from (the result of 1~m) (the result of sum and 1).
Set f[i][j] as the number of schemes with length I and module P as J.g=f[1], enumerate the remainder of the next module p, and you can transfer.
Perhaps you can double it. You can do it without crt+ntt.
#include <iostream> #include <fstream> #include <algorithm> #include <cmath> #include <ctime> #include <cstdio> #include <cstdlib> #include <cstring> using namespace std; #define mmst(a, b) memset(a, b, sizeof(a)) #define mmcp(a, b) memcpy(a, b, sizeof(b)) typedef long long LL; const int N=20020000,nn=2e7; const LL mo=20170408; int n,m,p; int prime[N],num; LL x[102],y[102],ans[102],by[102]; bool b[N]; void pre() { for(int i=2;i<=nn;i++) { if(!b[i]) prime[++num]=i; for(int j=1;j<=num&&i*prime[j]<=nn;j++) { b[prime[j]*i]=1; if(i%prime[j]==0) break; } } b[1]=1; } void cheng(LL *a,LL *b) { mmst(by,0); for(int i=0;i<p;i++) for(int j=0;j<p;j++) by[(i+j)%p]=(by[(i+j)%p]+a[i]*b[j])%mo; mmcp(a,by); } LL work(LL *a,int x) { mmst(ans,0); ans[0]=1; for(;x;x>>=1,cheng(a,a)) if(x&1) cheng(ans,a); return ans[0]; } int main() { pre(); cin>>n>>m>>p; for(int i=1;i<=m;i++) { x[i%p]++; if(b[i]) y[i%p]++; } cout<<(work(x,n)-work(y,n)+mo)%mo<<endl; return 0; }
This is the water theme specialty of yarn fog.