Bzoj 2038 small Z socks (Mo team algorithm)

MO Team Algorithm
If we know the answer to [l,r], we can get [L+1,r] 's answer in O (1) and the answer of [l,r-1], we can use the MO team algorithm. The time complexity is O (n^1.5). If the interval can only be moved at logn time, the time complexity is O (n^1.5*log N).
is to find a data structure to support the insertion, deletion, maintenance of the current answer.

It is easy to use an array to implement this problem, and to do O (1) transfer from [l,r] to [l,r+1] and [L+1,r].

So how does the MO team algorithm do? The following are the time complexities discussed on the basis of transfer to O (1). Also, because N and M are in the same order, write n uniformly.
If the answer to [L,r] is known, and the answer to [l ', R '] is required, we can easily get through the |l–l ' |+|r–r ' | sub-transfer.
Divides the number of n into sqrt (n) blocks.
Sort by interval, with the first keyword within the block of the left endpoint, and the right end as the second keyword, sorted by (POS [l],r)
And then by this sort of direct violence, the complexity analysis is like this:
1, I and i+1 in the same piece, r monotonically increment, so R is O (n). Because of the n^0.5 block, this part of the time complexity is n^1.5.
2, I and i+1 across a piece, r up to change N, because of the n^0.5 block, so this part of the time complexity is n^1.5
3, I and i+1 in the same piece of L change not more than n^0.5, across a piece will not exceed n^0.5, ignore * *. Because there are M queries (and N peers), the time complexity is n^1.5
So it's O (n^1.5).

# include <cstdio># include<cstring># include<cstdlib># include<iostream># include<vector># include<queue># include<stack># include<map># include<Set># include<cmath># include<algorithm>using namespacestd;# define Lowbit (x) ( x)& (-x)) # define PI3.1415926535# define EPS 1e-9# define MOD1000000009# define INF1000000000# define MEM (A, B) memset (A, B,sizeof(a) # define for (I,a,n) for(intI=a; i<=n; ++i) # define FO (I,a,n) for(intI=a; i<n; ++i) # define bug puts ("H"); # define LCH P<<1, l,mid# define RCH p<<1|1, mid+1, r# define MP make_pair# define PB push_backtypedef pair<int,int>pii;typedef Vector<int>vi;# pragma comment (linker,"/stack:1024000000,1024000000") typedefLong LongLL;intScan () {intres=0, flag=0; Charch; if((Ch=getchar ()) = ='-') flag=1; Else if(ch>='0'&&ch<='9') res=ch-'0';  while((Ch=getchar ()) >='0'&&ch<='9') res=res*Ten+ (ch-'0'); returnflag?-res:res;}voidOut (inta) {if(a<0) {Putchar ('-'); a=-A;} if(a>=Ten) Out (A/Ten); Putchar (A%Ten+'0');}Const intn=50005;//Code begin ...structquery{intL, R, id;} Node[n]; ll GCD (ll A, ll b) {returnb==0? A:GCD (b,a%b);}structans{LL A, B; voidReduce () {LL d=gcd (A, b); a/=d; b/=D;}} Ans[n];intA[n], num[n], N, M, unit;BOOLComp (query A, query B) {if(A.l/unit!=b.l/unit)returna.l/unit<b.l/Unit; returna.r<B.R;}voidWork () {LL temp=0; MEM (num,0); intL=1, r=0; FO (i,0, M) {         while(r<Node[i]. R) {++r; temp-= (LL) num[a[r]]*Num[a[r]]; ++num[a[r]]; temp+= (LL) num[a[r]]*Num[a[r]]; }         while(r>Node[i]. R) {Temp-= (LL) num[a[r]]*num[a[r]]; --Num[a[r]]; Temp+ = (LL) num[a[r]]*num[a[r]]; --s; }         while(l<Node[i]. L) {Temp-= (LL) num[a[l]]*num[a[l]]; --Num[a[l]]; Temp+ = (LL) num[a[l]]*num[a[l]]; ++m; }         while(l>Node[i]. L) {--l; temp-= (LL) num[a[l]]*Num[a[l]]; ++num[a[l]]; temp+= (LL) num[a[l]]*Num[a[l]]; } ans[node[i].id].a=temp-(r-l+1); ANS[NODE[I].ID].B= (LL) (r-l+1) * (rm);    Ans[node[i].id].reduce (); }}intMain () {scanf ("%d%d",&n,&m); For (I,1, N) scanf ("%d", A +i); FO (i,0, m) scanf ("%d%d", &node[i]. L,&node[i]. R), node[i].id=i; Unit=(int) sqrt (n); Sort (Node,node+M,comp);    Work (); FO (i,0, m) printf ("%lld/%lld\n", ans[i].a,ans[i].b); return 0;}

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Bzoj 2038 small Z socks (Mo team algorithm)