hdu5212 Code MO Team algorithm

This problem requires some knowledge definition notation for the MO team algorithm.
    
        f(A,B)    
  
 
    
 Indicates the answer with the query interval a, B, and the Notation + represents the collection and uses the MO team algorithm we can calculate any
    
        f(a,a)    
  
 
    
 The value may be assumed
   
    A=[L1,R1],B=[L2,R2],C  =[R1+1,L2?1]
   Easy to know
   
    F  (A,B)=F  (A+B+C  ,A+B+C  )+F  (C  ,C  )?F  (A+C  ,A+C  )?F  (C  +B,C  +B)
   So a query is split into four queries that can be done with the MO team algorithm the total time complexity is
   
     
     o   ( m  s  q   r   T   (  n  )   )          
    
   

 
     
      (above is the official puzzle)  
      

 
     
      code:  
      

   
    #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include < Cmath> #include <map>using namespace std;typedef long long ll;const int N = 30000*4 + 10;int pos[n];struct pp{&nbsp ;   int l,r,id;    int ans;} P[n];int CMP (pp a,pp b) {    if (POS[A.L]==POS[B.L]) return A.R < b.r;    return A.L < B.L;} int CMP2 (pp a,pp b) {    return a.id < b.id;} struct que{    int l1,l2,r1,r2;} Q[n];int Block,n,k,m,num;int A[n],cnt[n],answer; LL x;map<ll,int> mm;void Update (int x,int v) {    int val = k-a[x];    if (val <= 0) return;     Answer + = cnt[val]*v;    Cnt[a[x]] + = V;} void Solve () {    int l,r;    answer = 0;    for (int i=1,l=1,r=0;i<=num;i++) {//by block update &nbs P       for (; r<p[i].r;r++)             Update (r+1,1);      &NB Sp for (; r>p[i].r;r--)   &nbsp         Update (R,-1);        for (; l<p[i].l;l++)         &NBS P   Update (L,-1);        for (; l>p[i].l;l--)             Update (L-1 , 1);        P[i].ans = answer;   }}int main () {    while (scanf ("%d", &n)!=eof) {        mm.clear ();        num = 0;        block = (int) sqrt ( N) +1;        for (int i=1;i<=n;i++) Pos[i] = I/block + 1;        scanf ("%d", &amp k);        memset (cnt,0,sizeof (CNT));        for (int i=1;i<=n;i++)   & nbsp         scanf ("%d", &a[i]);        scanf ("%d", &m);      &NB Sp for (int i=1;i<=m;i++) {            int l1,l2,r1,r2;          &NBS P scanf ("%d%d%d%d ", &AMP;L1,&AMP;R1,&AMP;L2,&AMP;R2);            Q[I].L1 = L1, q[i].r1 = r1, q[i].l2 = L2, Q[i].r2 = r2;            P[++NUM].L = L1, p[num].r = l2-1;      &NBSP ;      x = l1*40000+ (l2-1);            MM[X] = num;            P[num].id = num;            P[++NUM].L = r1+1, P[NUM].R = R2;             x = (r1+1) *40000+r2;            MM[X] = num;            P[num].id = num;            if (l2-1 >= r1+1) {        & nbsp       P[++NUM].L = r1+1, P[NUM].R = l2-1;                x = (r1+ 1) *40000+ (l2-1);                MM[X] = num;          &NB Sp   &NBSP; P[num].id = num;           }            P[++NUM].L = L1, p[n UM].R = R2;            x = l1*40000+r2;            mm[x] = Nu m;            p[num].id = num;       }        sort ( P+1,P+1+NUM,CMP);        solve ();        sort (P+1,P+1+NUM,CMP2);    & nbsp   for (int i=1;i<=m;i++) {            int l1,l2,r1,r2;        &NBS P   L1 = q[i].l1, L2 = q[i].l2, r1 = q[i].r1, r2 = q[i].r2;            LL ans = 0;&nb Sp           x = l1*40000+r2;            ans + p[mm[x]].ans;  &N Bsp         if (l2-1>=r1+1) {                x = (r1+1) *40000+ (L2 -1); &NBSp               ans + p[mm[x]].ans;           }  &NB Sp         x = (r1+1) *40000+r2;            ans-= p[mm[x]].ans;  &nbsp ;         x = l1*40000+ (l2-1);            ans-= p[mm[x]].ans;  &nbsp ;         printf ("%lld\n", ans);       }   }    return 0;}
   

hdu5212 Code mo algorithm