莫队模版

本来想能不学莫队就不学，但是这次比赛被莫队卡得好惨，于是学了一发。。

先切道模板题。。

最普通的莫队就是把序列分块，然后按照询问所在的块来排序，减少指针的移动次数。。

#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 n, m, k, a[N], cnt[N], ans, res[N];struct Query {    int l, r, id, block;    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 k){    cnt[a[k]] ++;    ans += 2 * cnt[a[k]] - 1;}void remove(int k){    cnt[a[k]] --;    ans += -2 * cnt[a[k]] - 1;}int main(){    n = read(), m = read(), k = read();    int t = (int)sqrt(n);    for(int i = 1; i <= n; i ++) a[i] = read();    for(int i = 1; i <= m; i ++){        query[i].l = read(), query[i].r = read();        query[i].id = i, query[i].block = (query[i].l - 1) / t + 1;    }    sort(query + 1, query + m + 1);    int l = 1, r = 0;    for(int i = 1; i <= m; i ++){        int curL = query[i].l, curR = query[i].r;        while(l < curL) remove(l ++);        while(r < curR) add(++ r);        while(l > curL) add(-- l);        while(r > curR) remove(r --);        res[query[i].id] = ans;    }    for(int i = 1; i <= m; i ++){        printf("%d\n", res[i]);    }    return 0;}