莫队，卡常数

题目地址

• 思路

  • 设\(\text{Vis[i]}\)为元素\(\text{i}\)在区间\(\text{[L,R]}\)的出现次数

  • 考虑区间\(\text{[L,R]}\)和元素\(\text{i}\)，首次取出的概率为\(\frac{Vis[i]}{R-L+1}\)，再次取出的概率是\(\frac{Vis[i]-1}{R-L}\)

  • 对于区间\(\text{[L,R]}\)，答案为\(\sum_{i=1}^{N}{\frac{Vis[i](Vis[i]-1)}{(R-L)*(R-L+1)}}\)。

  • 这样，每次给变\(\text{[L,R]}\)，分子的变化可以通过对前、后两项的值做差得到：

    • 设X=Vis[i],有：

    \[ (X+1)X-(X)(X-1)=2X \]

  • 莫队这样修改即可。

• Code

  • /**************************************************************Problem: 2038User: Bj2002Language: C++Result: AcceptedTime:11036 msMemory:2584 kb
    ****************************************************************/#include <stdio.h>
    #include <string.h>
    #include <algorithm>
    #define GC getchar()
    #define Clean(X,K) memset(X,K,sizeof(X))
    using namespace std ;
    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 ;
    }
    long long GCD(long long M, long long N) {while (N != 0) {long long T = M % N;M = N;N = T;}return M;
    }
    const int Maxn = 50005 ;
    int N , M , A[Maxn] , Vis[Maxn]  ;
    long long Ans[Maxn] , Sum[Maxn];
    struct Node {int Left , Right  , Place;
    };
    Node Q[Maxn] ;
    bool Cmp (const Node &X , const Node &Y) {if (X.Left != Y.Left ) return X.Left < Y.Left ;if (X.Left & 1) return X.Right < Y.Right ;else return X.Right > Y.Right ;
    }
    bool Cmp2 (const Node &X , const Node &Y) {return X.Place < Y.Place ;
    }
    void Qwrite(int X) {if(X > 9) Qwrite(X / 10);putchar(X % 10 + '0');
    }int main () {
    //  freopen ("P1494.in" , "r" , stdin) ;
    //  freopen ("P1494.out", "w" , stdout) ;N = Qread () , M = Qread ();for (int i = 1 ; i <= N; ++ i) A[i] = Qread () ;for (int i = 1 ; i <= M; ++ i) Q[i].Left = Qread () , Q[i].Right = Qread () , Q[i].Place = i ;;sort (Q + 1 , Q + 1 + M , Cmp) ;Clean (Vis , 0) ;int L , R ;long long Now = 0 ;L = R = Q[1].Left ;Vis[A[L]] = 1 ;for (int i = 1 ; i <= M; ++ i) {while (L < Q[i].Left ) {-- Vis[A[L]] ;Now -= (Vis[A[L]] << 1) ;++ L ;}
    //      cout << L <<' '<<R <<' '<<Now<<endl;while (R < Q[i].Right ) {++ R ;Now += (Vis[A[R]] << 1) ;++ Vis[A[R]] ;}
    //      cout << L <<' '<<R <<' '<<Now<<endl;while (R > Q[i].Right ) {-- Vis[A[R]] ;Now -= (Vis[A[R]] << 1) ;-- R ;}
    //      cout << L <<' '<<R <<' '<<Now<<endl;Ans[Q[i].Place] = Now , Sum[Q[i].Place ] = ((long long)Q[i].Right - Q[i].Left ) * (Q[i].Right - Q[i].Left + 1) ;}std :: sort (Q + 1 , Q + 1 + M , Cmp2) ;for (int i = 1 ; i <= M; ++ i) {if (Q[i].Left == Q[i].Right ) printf ("0/1\n") ;else {int K = GCD (Ans[i] , Sum[i]) ;Ans[i] /= K , Sum[i] /= K ;printf ("%lld/%lld\n" , Ans[i] , Sum[i]) ;}}fclose (stdin) , fclose (stdout) ;return 0 ;
    }

Thanks!