[题解] 2038: [2009国家集训队]小Z的袜子(hose)

莫队，卡常数

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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: 2038
        User: Bj2002
        Language: C++
        Result: Accepted
        Time:11036 ms
        Memory: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!