又是一道奶牛题
从左到右扫描,树状数组维护【左端点出现而右端点未出现】的数字的个数。记录每个数字第一次出现的位置。
若是第二次出现,那么删除第一次的影响。
#include <cstdio> #include <cstring> #define re register #define GC getchar() #define Lowbit(X) (X&(-X)) #define Clean(X,K) memset(X,K,sizeof(X)) int Qread () { int X = 0 ; char C = GC ; while (C > '9' || C < '0') C = GC ; while (C >='0' && C <='9') {X = X * 10 + C - '0';C = GC ;} return X ; } const int Maxn = 50005 << 1 ; int N , Vis[Maxn >> 1] , A[Maxn] , Head[Maxn >> 1] , T[Maxn]; unsigned long long int Ans = 0 ; int Ask (int X) {re int Ans = 0 ;while (X > 0) { Ans += T[X] ; X -= Lowbit(X) ;}return Ans ;} void Add (int X , int K) {while (X <= N) {T[X] += K;X += Lowbit(X) ;}} int main () { N = Qread () * 2 , Clean(Vis , 0) ; for (re int i = 1 ;i <= N; ++ i) A[i] = Qread () ; for (re int i = 1 ;i <= N; ++ i)if (Vis[A[i]]) Add (Head[A[i]] , -1) , Ans += Ask (i) - Ask (Head[A[i]]) ; else Head[A[i]] = i , Add (i , 1) , Vis[A[i]] = 1 ; printf ("%lld " , Ans) ; fclose (stdin) ,fclose (stdout); return 0; }