BZOJ2038[2009 Country Training Team] small Z socks (hose)

BZOJ2038[2009 Country Training Team] small Z socks (hose)

Test instructions

The n socks are numbered from 1 to N, and each time you draw two from the socks numbered L to R, how much is the probability of pumping the same socks with the same color.

Exercises

Do not know what data structure to use, but can use a global array to save the current number of each color, so that by the interval [l,r] to launch [l,r±1] the answer and [l±1,r] the complexity of O (1), for this problem, can be the complexity of O (NSQRT (n)) of the MO team algorithm to solve.

The MO team algorithm is an off-line algorithm that sorts the queries in some order, making the averaging complexity O (nsqrt (n)), how to sort? If you sort by the left endpoint, then R will be possible to swing a large number of times, so that the complexity is degraded to O (n^2), the positive solution is the end of the block, so that the left end of the block as the first keyword sort, the right end of the second keyword sort. This way, when two queries L are in the same piece, L can only move sqrt (n) times. L in the same block of multiple queries Q can only shift n times, L in different blocks R may shift n times, but because only sqrt (n) block, so the operation to move N is only sqrt (n) times, so the averaging complexity is O (sqrt (n)). All the averaging complexity is metaphysics ...

Because the intermediate results are not forced into a long long,wa several hair, the weak is too weak!

Code:

1#include <cstdio>2#include <cstring>3#include <algorithm>4#include <cmath>5 #definell Long Long6 #defineInc (I,J,K) for (int i=j;i<=k;i++)7 using namespacestd;8 9 structnd1{Ten     intL,pl,r,id; ll ans; One }; A BOOLCmp1 (nd1 a,nd1 b) { -     if(a.pl!=b.pl)returna.pl<b.pl;if(A.R!=B.R)returna.r<B.R; -     returna.l<B.L; the } - BOOLCMP2 (nd1 a,nd1 b) { -     returna.id<b.id; - } +Nd1 a1[100000];intcol[100000],pos[100000],n,m,l,r;ll ans,s[100000]; -InlinevoidUpdateintXinty) { +ans-= (s[col[x]]* (s[col[x]]-1)); s[col[x]]+= (LL) y;ans+= (s[col[x]]* (s[col[x]]-1)); A } atll GCD (ll A,ll b) {returnb==0? A:GCD (b,a%b);} - intMain () { -scanf"%d%d", &n,&m); Inc (I,1, N) scanf ("%d", &col[i]);intSz= (int) sqrt (n); -Inc (I,1, n) pos[i]= (i-1)/sz+1; -Inc (I,1, M) { -         intA,B;SCANF ("%d%d", &a,&b); a1[i]= (nd1) {a,pos[a],b,i,0}; in     } -Sort (a1+1, a1+1+M,CMP1); L=1; R=0; ans=0; memset (s),0,sizeof(s)); toInc (I,1, M) { +          while(R&LT;A1[I].R) Update (r+1,1), r++; -          while(L&GT;A1[I].L) Update (L-1,1), l--; the          while(R&GT;A1[I].R) Update (r,-1), r--; *          while(L&LT;A1[I].L) Update (l,-1), l++; $a1[i].ans=ans;Panax Notoginseng     } -Sort (a1+1, a1+1+M,CMP2);  theInc (I,1, M) { +         if(a1[i].ans==0) printf ("0/1\n");Else{ All A2=GCD (A1[i].ans, (LL) (a1[i].r-a1[i].l+1) * (a1[i].r-a1[i].l)); theprintf"%lld/%lld\n", A1[I].ANS/A2, (LL) (a1[i].r-a1[i].l+1) * (A1[I].R-A1[I].L)/A2); +         } -     } $     return 0; $}

20160405

BZOJ2038[2009 Country Training Team] small Z socks (hose)