比较套路的一个题, 对每个数维护一颗线段树来转移就好了.
#include <iostream> #include <algorithm> #include <cstdio> #include <math.h> #include <set> #include <map> #include <queue> #include <string> #include <string.h> #include <bitset> #define REP(i,a,n) for(int i=a;i<=n;++i) #define PER(i,a,n) for(int i=n;i>=a;--i) #define hr putchar(10) #define pb push_back #define lc tr[o].l #define rc tr[o].r #define mid ((l+r)>>1) #define ls lc,l,mid #define rs rc,mid+1,r #define x first #define y second #define io std::ios::sync_with_stdio(false) #define endl '\n' using namespace std; typedef long long ll; typedef pair<int,int> pii; const int P = 1e9+7, INF = 0x3f3f3f3f; ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;} ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;} ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;} //head const int N = 1e5+10; int n, m, ans, tot, T[N]; int a[N], s[N], b[N], c[N], L[N], R[N]; int sub(int x, int y) {int r=x-y;if(r<0)r+=P;return r;} int add(int x, int y) {int r=x+y;if(r>=P)r-=P;return r;} int mul(int x, int y) {return (ll)x*y%P;} struct _ { int A,C,S,AB,BC,ABC; _ () {} _ (int A, int C, int S) : A(A),C(C),S(S),AB(0),BC(0),ABC(0) {} _ operator + (const _ &rhs) const { _ r; r.A = add(A,rhs.A); r.C = add(C,rhs.C); r.S = add(S,rhs.S); r.AB = add(add(AB,rhs.AB),mul(A,rhs.S)); r.BC = add(add(BC,rhs.BC),mul(S,rhs.C)); r.ABC = add(add(ABC,rhs.ABC),add(mul(AB,rhs.C),mul(A,rhs.BC))); return r; } }; struct {int l,r;_ v;} tr[N<<5]; void ins(int &o, int l, int r, int x, int v1, int v2, int v3) { if (!o) o=++tot; if (l==r) { tr[o].v=_(v1,v2,v3),void(); return ; } if (mid>=x) ins(ls,x,v1,v2,v3); else ins(rs,x,v1,v2,v3); tr[o].v=tr[lc].v+tr[rc].v; } int main() { scanf("%d", &n); REP(i,1,n) scanf("%d", a+i),b[i]=a[i]; sort(b+1,b+1+n),*b=unique(b+1,b+1+n)-b-1; REP(i,1,n) { a[i] = lower_bound(b+1,b+1+*b,a[i])-b; for (int j=a[i]; j; j^=j&-j) L[i]+=c[j]; for (int j=a[i]; j<=*b; j+=j&-j) ++c[j]; } memset(c,0,sizeof c); PER(i,1,n) { for (int j=a[i]; j; j^=j&-j) R[i]+=c[j]; for (int j=a[i]; j<=*b; j+=j&-j) ++c[j]; } REP(i,1,n) { ans = sub(ans,tr[T[a[i]]].v.ABC); ins(T[a[i]],1,n,i,L[i],R[i],1); ans = add(ans,tr[T[a[i]]].v.ABC); } scanf("%d", &m); REP(i,1,m) { int op, x; scanf("%d%d", &op, &x); ans = sub(ans,tr[T[a[x]]].v.ABC); ins(T[a[x]],1,n,x,(op==2)*L[x],(op==2)*R[x],op==2); ans = add(ans,tr[T[a[x]]].v.ABC); printf("%d\n", ans); } }