BZOJ3745: [Coci2015]norma

Description

Input line 1th, an integer n, line 2~n+1, an integer per line representing sequence a.

Output outputs the result of the answer to the 10^9 modulo.

Sample Input4
2
4
1
4Sample Output109

"Data Range"
N <= 500000
1 <= a_i <= 10^8HINT Source

Acknowledgement Dzy

That's ********.

Considering the divide-and-conquer approach, the question turns into how to solve the cross-midpoint answer.

For a sequence [i,j], the reverse enumeration of the left endpoint I, the entire value of MAXV, MINV, there are four possible:

1. All are taken at the left half.

Calculate the right end to meet the conditions up to where, remember as cur, because of monotonicity, can O (1) to beg.

ans+=maxvl*minvl* ((mid+1-i+1) + (mid+2-i+1) + + (cur-i+1))

Deformation, with arithmetic progression sum can O (1) Forget it.

2.MAXV is taken on the left, MINV in the right.

Calculate the right end to meet the conditions up to where, remember as CURMN, because of monotonicity, can O (1) to beg.

ans+=maxvl*[minvj* (j-i+1) +minvj+1* (j+1-i+1) + +minvcurmn* (curmn-i+1)]

Deformation, maintenance of minvj*j and MINVJ prefix and can O (1) Forget it.

3.MINV is MAXV on the right, and the left is taken.

Calculate the right end to meet the conditions up to where, remember as CURMX, because of monotonicity, can O (1) to beg.

ans+=minvl*[maxvj* (j-i+1) +maxvj+1* (j+1-i+1) + +maxvcurmx* (curmx-i+1)]

Deformation, maintenance of maxvj*j and MAXVJ prefix and can O (1) Forget it.

4. All are taken in the right half.

ans+=maxvj*minvj* (j-i+1) + +maxvr*minvr* (r-i+1).

Deformation, maintenance of maxvj*minvj*j and MAXVJ*MINVJ prefix and can O (1) Forget it.

It's really easy to say!

#include <cstdio>#include<cctype>#include<queue>#include<cmath>#include<cstring>#include<algorithm>#defineRep (i,s,t) for (int i=s;i<=t;i++)#defineDwn (i,s,t) for (int i=s;i>=t;i--)#defineren for (int i=first[x];i!=-1;i=next[i])using namespacestd;Const intBuffersize=1<< -;Charbuffer[buffersize],*head,*Tail;inlineCharGetchar () {if(head==tail) {        intL=fread (Buffer,1, Buffersize,stdin); Tail= (Head=buffer) +l; }    return*head++;} InlineintRead () {intx=0, f=1;CharC=Getchar ();  for(;! IsDigit (c); C=getchar ())if(c=='-') f=-1;  for(; IsDigit (c); C=getchar ()) x=x*Ten+c-'0'; returnx*F;}Const intmaxn=500010;Const intMod=1000000000; typedefLong Longll;structPair {ll mx,mn;} Q1[MAXN],Q2[MAXN]; Pair update (pair A,ll v) {a.mx=max (a.mx,v); a.mn=min (a.mn,v); returnA;}intn;ll a[maxn],ans,s1[maxn],s2[maxn],s3[maxn],ss1[maxn],ss2[maxn],ss3[maxn];ll Sumlen (intBintSintt) {if(s>t)return 0; return(LL) (s-b+1+t-b+1) * (t-s+1)/2%MoD;}voidSolveintLintr) {if(r-l+1<=Ten) {Rep (i,l,r) {ll MX=a[i],mn=A[i]; Rep (j,i,r) {MX=max (Mx,a[j]); mn=min (mn,a[j]); (Ans+ = (j-i+1) *mx*mn)%=MoD; }        }        return; }    intMid=l+r>>1; Solve (l,mid); Solve (mid+1, R); Q1[mid]= (Pair) {A[mid],a[mid]};d WN (i,mid-1, l) q1[i]=update (q1[i+1],a[i]); Q2[mid+1]= (Pair) {a[mid+1],a[mid+1]};rep (i,mid+2, R) q2[i]=update (q2[i-1],a[i]); Q1[l-1]=q2[r+1]= (Pair) {1e9,-1e9}; S1[mid]=s2[mid]=s3[mid]=ss1[mid]=ss2[mid]=ss3[mid]=0; Rep (I,mid+1, R) {S1[i]= (s1[i-1]+q2[i].mx* (i-mid))%MoD; Ss1[i]= (ss1[i-1]+q2[i].mx)%MoD; S2[i]= (s2[i-1]+q2[i].mn* (i-mid))%MoD; Ss2[i]= (ss2[i-1]+Q2[I].MN)%MoD; S3[i]= (s3[i-1]+q2[i].mx*q2[i].mn%mod* (i-mid))%MoD; Ss3[i]= (ss3[i-1]+q2[i].mn*q2[i].mx)%MoD; }    intCur=mid+1, curmx=mid+1, curmn=mid+1; Dwn (i,mid,l) { while(q2[cur].mx<=q1[i].mx&&q2[cur].mn>=q1[i].mn) cur++;  while(q2[curmx].mx<=q1[i].mx) curmx++;  while(q2[curmn].mn>=q1[i].mn) curmn++; ll T=ans; (Ans+=sumlen (i,mid+1, cur-1) *q1[i].mx%mod*q1[i].mn%mod)%=MoD; if(curmn-1&GT;=CURMX) (ans+=q1[i].mn* (s1[curmn-1]-s1[curmx-1]+mod+ (ss1[curmn-1]-ss1[curmx-1]+MOD) * (mid-i+1)) (%mod))%=MoD; if(curmx-1&GT;=CURMN) (ans+=q1[i].mx* (s2[curmx-1]-s2[curmn-1]+mod+ (ss2[curmx-1]-ss2[curmn-1]+MOD) * (mid-i+1)) (%mod))%=MoD; intp=Max (CURMX,CURMN); if(P&LT;=R) (ans+=s3[r]-s3[p-1]+mod+ (ss3[r]-ss3[p-1]+MOD) * (mid-i+1))%=MoD; }}intMain () {n=read (); Rep (I,1, N) a[i]=read (); Solve (1, n);p rintf ("%lld\n", ans); return 0;}

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BZOJ3745: [Coci2015]norma