20181009noip HZ EZ Two-school joint test sum (Mo Team, combinatorial mathematics)

Face stamp Here

Idea: Noip team??!!!

The examination room is dead or alive not to this aspect to think AH!!! Data divide forget write endl50pts roll thick

Every query here has n,m two parameters

We can think of it as the L and R in the regular team.

Then we can use the recursive nature of the combined number.

Believe in Change m everyone will write, n?

Look at the picture:

We found that $S _n^m = s_{n-1}^m \times 2-c_n^{m+1} + c_{n-1}^{m+1}$

(Because of the nature of the Yang Hui triangle)

So n can also be recursive

Just set up a MO team.

Code:

#include <iostream>#include<cstdio>#include<algorithm>#defineRii Register int i#defineRij Register Int J#defineint long LongConst intp=1e9+7;using namespacestd;structque{intN,M,ANS,BH;} x[100005];intny[100005],jc[100005],jcny[100005],q,sy[100005],nn,nm,ans;voidQny () {ny[1]=1; ny[0]=1;  for(rii=2; i<=100000; i++) {Ny[i]= (p-p/i) *ny[p%i]%p; }}voidJic () {jcny[0]=1; jc[1]=1; jcny[1]=1;  for(rii=2; i<=100000; i++) {Jc[i]=jc[i-1]*i; Jc[i]%=p; Jcny[i]=jcny[i-1]*Ny[i]; Jcny[i]%=p; }}voidYcl () {qny (); JIC ();}BOOLcmp1 (que lk,que KL) {if(sy[lk.n]==SY[KL.N]) {        returnlk.m<kl.m; }    returnlk.n<KL.N;}BOOLcmp2 (que lk,que KL) {returnlk.bh<kl.bh;}intCintNintm) {    intkkk=Jcny[m]; KKK*=Jc[n]; KKK%=p; KKK*=jcny[n-m]; KKK%=p; returnKKK;}voidcnintZL) {    if(zl==1) {ans*=2; Ans-=c (nn+1, nm+1); Ans+=c (nn,nm+1); Ans+=p; Ans%=p; }    if(zl==-1) {ans+=c (nn,nm+1); Ans-=c (nn-1, nm+1); Ans+=p; Ans%=p; Ans*=ny[2]; Ans%=p; }}voidcmintZL) {    if(zl==1) {ans+=c (nn,nm+1); }    Else{ans-=c (NN,NM); Ans+=p; } ans%=p;} Signed Main () {Freopen ("sum.in","R", stdin); Freopen ("Sum.out","W", stdout);    YCL (); scanf ("%lld",&q); scanf ("%lld",&q); intlen= the;  for(rii=1; i<=100000; i++) {Sy[i]=i/len+1; }     for(rii=1; i<=q;i++) {scanf ("%lld%lld",&x[i].n,&x[i].m); X[I].BH=i; } sort (x+1, x+q+1, CMP1); NN=1, nm=0, ans=1;  for(rii=1; i<=q;i++)    {        if(x[i].m<nn) {             while(nm>x[i].m) {cm (-1); NM--; }             while(nm<x[i].m) {cm (1); NM++; }             while(nn<X[I].N) {cn (1); NN++; }             while(nn>X[I].N) {cn (-1); NN--; }        }         while(nn<X[I].N) {cn (1); NN++; }         while(nn>X[I].N) {cn (-1); NN--; }         while(nm>x[i].m) {cm (-1); NM--; }         while(nm<x[i].m) {cm (1); NM++; } X[i].ans=ans; } sort (x+1, x+q+1, CMP2);  for(rii=1; i<=q;i++) {printf ("%lld\n", X[i].ans); }}

20181009noip HZ EZ Two-school joint test sum (Mo Team, combinatorial mathematics)