莫队 + 树状数组

先离散化，把值域离散出来。

每次莫队指针移动的时候，用树状数组维护该范围的值域即可。

具体来说，如果有一个数在左边加入（删除），就用树状数组查询比他小的数，右边同理，查询比他大的数就行了。和逆序对差不多的感觉。

#include 
    
     #define INF 0x3f3f3f3f#define full(a, b) memset(a, b, sizeof a)using namespace std;typedef long long ll;inline int lowbit(int x){ return x & (-x); }inline int read(){    int X = 0, w = 0; char ch = 0;    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();    return w ? -X : X;}inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }inline int lcm(int a, int b){ return a / gcd(a, b) * b; }template
     
      inline T max(T x, T y, T z){ return max(max(x, y), z); }template
      
       inline T min(T x, T y, T z){ return min(min(x, y), z); }template
       
        inline A fpow(A x, B p, C lyd){    A ans = 1;    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;    return ans;}const int N = 50005;int t[N], a[N], b[N], n, q, k, s, ans, res[N];struct Query{    int l, r, block, id;    bool operator < (const Query &rhs) const {        return (block ^ rhs.block) ? l < rhs.l : (block & 1) ? r < rhs.r : r > rhs.r;    }}query[N];void add(int index, int i){    for(; index <= k; index += lowbit(index))        t[index] += i;}int response(int index){    int ret = 0;    for(; index; index -= lowbit(index))        ret += t[index];    return ret;}int main(){    n = read();    for(int i = 1; i <= n; i ++) b[i] = a[i] = read();    sort(b + 1, b + n + 1);    k = (int)(unique(b + 1, b + n + 1) - b - 1);    for(int i = 1; i <= n; i ++){        a[i] = (int)(lower_bound(b + 1, b + k + 1, a[i]) - b);    }    q = read(), s = (int)sqrt(n);    for(int i = 1; i <= q; i ++){        query[i].l = read(), query[i].r = read();        query[i].id = i, query[i].block = (query[i].l - 1) / s + 1;    }    int l = 1, r = 0;    sort(query + 1, query + q + 1);    for(int i = 1; i <= q; i ++){        int curL = query[i].l, curR = query[i].r;        while(l < curL) add(a[l], -1), ans -= response(a[l] - 1), l ++;        while(r < curR) r ++, add(a[r], 1), ans += r - l + 1 - response(a[r]);        while(l > curL) l --, add(a[l], 1), ans += response(a[l] - 1);        while(r > curR) add(a[r], -1), ans -= r - l - response(a[r]), r --;        res[query[i].id] = ans;    }    for(int i = 1; i <= q; i ++){        printf("%d\n", res[i]);    }    return 0;}