Sum of the products of 51nod 1348

The sum of all the schemes (0<=i<=r-l+1) multiplied by the number of f[l,r in the interval [l,r] is represented by the (r-l+2) dimension vector], which can be found f[l,r]=f[l,m]*f[m+1,r], the topic modulus 100003 is smaller, The upper bound of the answer is approximately 1e16 after each convolution, the CRT can be combined with NTT in the two-1e9 modulus, the complexity is O (nlog2n), and the constant optimization should be noted.

#include <cstdio>#include<cstring>#include<algorithm>typedefLong Longi64;int _(){    intx=0, f=1, c=GetChar ();  while(c< -) c=='-'&& (f=-1), c=GetChar ();  while(c> -) x=x*Ten+c- -, c=GetChar (); returnx*F;}intPowintAintNintp) {    intv=1;  for(; n;n>>=1){        if(n&1) v=i64 (v) *a%p; A=i64 (a) *a%p; }    returnv;} I64 Mul (i64 a,i64 b,i64 p) {i64 s=0; A%=p;b%=p;  while(b) {if(b&1) (S+=a)%=p; (A<<=1)%=p; b>>=1; }    returns;}intn,x,r[262144];template<Const intPConst intG>voidNttint*a,intt) {     for(intI=0; i<n;++i)if(i>R[i]) Std::swap (A[i],a[r[i]);  for(intI=1; i<n;i<<=1){        intW=pow (g, (t* (P-1)/(i*2) +p-1), p);  for(intj=0;j<n;j+=i<<1){            intE=1, *b=a+j,*c=b+i;  for(intk=0; k<i;++k,e=i64 (e) *w%p) {                intX=b[k],y=c[k]*i64 (e)%p; B[K]= (x+y)%p; C[K]= (x-y+p)%p; }        }    }    if(t==-1) {i64 I=pow (n,p-2, p);  for(intI=0; i<n;++i) a[i]=a[i]*i%p; }}intn,q,v0[50007],mem[65536* -+7],*mp=mem,vs[4][65536+7];Const intp1=998244353, g1=3, p2=950009857, g2=7;ConstI64 m1=i64 (p1) *pow (p1,p2-2, p2), m2=i64 (p2) *pow (p2,p1-2, p1), ps=i64 (p1) *P2;int*calc (intLintR) {    int*pos=mp;mp+=r-l+1; if(l==R) {        *pos=V0[l]; returnPOS; }    intM=l+r>>1; int*lp=calc (l,m)-1; int*rp=calc (m+1, R)-1;  for(n=2, x=0; n<r-l+2; n<<=1,++X); if(r-l+1<= -){         for(intI=0;i<2; ++i) memset (Vs[i],0Nsizeof(int)), vs[i][0]=1;  for(intI=1; i<=m-l+1; ++i) vs[0][i]=Lp[i];  for(intI=1; i<=r-m;++i) vs[1][i]=Rp[i];  for(intI=1; i<=r-l+1;++i) {             for(intj=0; j<=i;++j) pos[i-1]= (pos[i-1]+i64 (vs[0][J]) *vs[1][I-J])%100003; }        returnPOS; }     for(intI=1; i<n;++i) r[i]=r[i>>1]>>1| (i&1) <<y;  for(intI=0;i<4; ++i) memset (Vs[i],0Nsizeof(int)), vs[i][0]=1;  for(intI=1; i<=m-l+1; ++i) vs[0][i]=vs[1][i]=Lp[i];  for(intI=1; i<=r-m;++i) vs[2][i]=vs[3][i]=Rp[i]; NTT<p1,g1> (vs[0],1); NTT<p1,g1> (vs[2],1); NTT<p2,g2> (vs[1],1); NTT<p2,g2> (vs[3],1);  for(intI=0; i<n;++i) vs[0][i]=i64 (vs[0][i]) *vs[2][i]%P1;  for(intI=0; i<n;++i) vs[1][i]=i64 (vs[1][i]) *vs[3][i]%P2; NTT<p1,g1> (vs[0],-1); NTT<p2,g2> (vs[1],-1);  for(intI=1; i<=r-l+1; ++i) pos[i-1]= (Mul (m2,vs[0][I],PS) +mul (m1,vs[1][I],PS))%ps%100003; returnPos;}intMain () {n=_ (); q=_();  for(intI=1; i<=n;++i) v0[i]=_(); int*ans=calc (1, N)-1;  while(q--) printf ("%d\n", Ans[_ ()]); return 0;}

Sum of the products of 51nod 1348