Bzoj thousand plan 309:bzoj4332:jsoi2012 snacks (divide and conquer FFT)

https://www.lydsy.com/JudgeOnline/problem.php?id=4332

Because if a child does not get candy, then the children behind her will not get candy.

So set g[i][j] to show the former I children, to the J Candy, and the former I children are divided into the program number of sweets

So that f (x) indicates the degree of happiness of x candies.

∴g[i][j] =∑g[i-1][j-k]*f (k)

Kee G[i]=g[i-1]*f, then g[i]=f ^ i

But the requirements are σg[i][m]

Kee F[n]=σg[i] i∈[1,n], then ans=f[n][m]

F[n]=σg[i] I∈[1,n]

=σf (N/2) +σg[i] i∈[n/2+1,n]

=σf (N/2) +σf^i I∈[n/2+1,n]

=σf (N/2) +σf^ (n/2+i) I∈[1,N/2]

=σf (N/2) +f^ (N/2) *σf^i I∈[1,N/2]

=σf (N/2) +g (N/2) *f (N/2)

Then you can divide and conquer the FFT solution

If n is odd, f (n) =f (n-1) +g[n]=f (n-1) +g (n-1) *f

Boundary Condition: g[][0]=1

#include <cmath>#include<cstdio>#include<algorithm>using namespacestd;Const intm=1<< -;#defineN 10001intM,mod;intr[m+1];intLen;Const DoublePi=acos (-1);structcomplex{Doublex, y; Complex () {} Complex (DoubleX_Doubley_): X (x_), Y (y_) {} Complexoperator+(Complex p) {Complex C; c.x=x+p.x; C.y=y+p.y; returnC; } Complexoperator-(Complex p) {Complex C; c.x=x-p.x; C.y=y-p.y; returnC; } Complexoperator*(Complex p) {Complex C; c.x=x*p.x-y*p.y; C.y=x*p.y+y*p.x; returnC; }    voidClear () {x=y=0; }};typedef Complex E; E f[m+1],f[m+1],g[m+1],tmp[m+1];voidFFT (E *a,intty) {     for(intI=0; i<len;++i)if(i<R[i]) swap (a[i],a[r[i]);  for(intI=1; i<len;i<<=1) {E wn (cos (pi/i), Ty*sin (pi/i));  for(intp=i<<1, j=0; j<len;j+=p) {E W (1,0);  for(intk=0; k<i;++k,w=w*WN) {E x=a[j+k],y=w*a[i+j+K]; A[j+k]=x+y; a[i+j+k]=x-y; }        }    }    if(ty==-1)    {         for(intI=0; i<len;++i) a[i].x=a[i].x/len,a[i].x=int(a[i].x+0.5)%mod,a[i].y=0; }}voidSolve (E *f,e *g,intN) {    if(!N) {g[0].x=1; return; }    if(n&1) {Solve (F,g,n-1); FFT (g,1);  for(intI=0; i<len;++i) g[i]=g[i]*F[i]; FFT (g,-1);  for(intI=0; i<=m;++i) f[i]=f[i]+G[i];  for(intI=0; i<=m;++i) f[i].x=int(f[i].x)%mod,f[i].y=0;  for(inti=m+1; i<len;++i) f[i].clear (), g[i].clear (); }    Else{Solve (f,g,n/2);  for(intI=0; i<len;++i) tmp[i]=F[i]; FFT (TMP,1); FFT (g,1);  for(intI=0; i<len;++i) tmp[i]=tmp[i]*G[i]; FFT (TMP,-1);  for(intI=0; i<len;++i) g[i]=g[i]*G[i]; FFT (g,-1);  for(intI=0; i<=m;++i) f[i]=f[i]+Tmp[i];  for(intI=0; i<=m;++i) f[i].x=int(f[i].x)%mod,f[i].y=0;  for(inti=m+1; i<len;++i) f[i].clear (), g[i].clear (); }}intMain () {intN,o,s,u; scanf ("%d%d%d%d%d%d",&m,&mod,&n,&o,&s,&u); //f[0].x=1;     for(intI=1; i<=m;++i) f[i].x= (o*i*i+s*i+u)%MoD; intL=0;  for(len=1; len<=m+m;len<<=1, l++);  for(intI=0; i<len;++i) r[i]= (r[i>>1]>>1)| ((i&1) <<l-1); FFT (F,1);    Solve (f,g,n); printf ("%d",int(f[m].x)); return 0;}

Bzoj thousand plan 309:bzoj4332:jsoi2012 snacks (divide and conquer FFT)