莫队算法模板

洛谷模板题

#include<bits/stdc++.h>
#include<unordered_map>
#include<math.h>
#include<forward_list>
#include<unordered_set>
using namespace std;
#pragma warning(disable:4996)
const double pi = acos(-1);
#define pdd pair<double,double>
#define re register
#define ll long long
#define endl '\n'
#define pii pair<ll,ll>
#define IOS std::ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fi first
#define se second
#define ull unsigned ll
#define eb emplace_back
#define pb push_back
#define power qpow
#define rad read
#define mp make_pair
#define ls (o<<1)
#define rs (o<<1|1)
#define mid (l+r>>1)
#define fpn(x) fp(x),putchar('\n')
ll mod = 1e9 + 7;
inline ll lowbit(ll x) { return x & (-x); }
inline ll  gcd(ll a, ll b) {
	return b ? gcd(b, a % b) : a;
}
inline ll exgcd(ll a, ll b, ll& x, ll& y) {
	//ax+by=1; 返回值为a,b最大公约数
	if (!b) {
		x = 1, y = 0;
		return a;
	}
	ll g = exgcd(b, a % b, y, x);
	y -= (a / b) * x;
	return g;
}
inline ll ksc(ll a, ll b, ll mod) {
	ll ans = 0;
	while (b) {
		if (b & 1) ans = (ans + a) % mod;
		a = (a + a) % mod;
		b >>= 1;
	}
	return ans;
}
ll qpow(ll a, ll b, ll modd = mod) {
	ll ans = 1;
	while (b) {
		if (b & 1)
			ans = ksc(ans, a, modd);
		a = ksc(a, a, modd);
		b >>= 1;
	}
	return ans;

}
inline ll read() {
	ll s = 0; bool w = true;
	char c = getchar();
	for (; c < '0' || c>'9'; c = getchar())if (c == '-')w = false;
	for (; c >= '0' && c <= '9'; c = getchar())s = (s << 1) + (s << 3) + (c ^ 48);
	return w ? s : -s;
}
inline void fp(ll n) {
	if (n < 0)  putchar('-'), n = -n;
	if (n > 9) fp(n / 10);
	putchar(n % 10 + 48);
}
const int N = 1e5 + 10;
struct node {
	int l, r, id, pos;
	//l,r,左右边界，id为询问编号，pos为分块所在位置
	bool operator<(const node& x) {
		if (pos == x.pos)return r < x.r;
		return pos < x.pos;
	}
}a[N];
int b[N];
ll cnt[N], mm[N];
void solve() {
	ll n = read(), m = read(), k = read();
	ll size = sqrt(n);
	for (int i = 0; i < n; ++i)b[i] = read();
	for (int i = 0; i < m; ++i) {
		a[i].l = read()-1;
		a[i].r = read()-1;
		a[i].id = i;
		a[i].pos = (a[i].l - 1) / size + 1;
	}
	sort(a, a + m);
	ll l = 0, r = -1,sum=0;
	//双指针移动
	for (int i = 0; i < m; ++i) {
		while (l > a[i].l)--l, cnt[b[l]]++, sum += 2 * cnt[b[l]] - 1;
		while (r < a[i].r)++r, cnt[b[r]]++, sum += 2 * cnt[b[r]] - 1;
		while (l < a[i].l)cnt[b[l]]--, sum -= 2 * cnt[b[l]] + 1, ++l;
		while (r > a[i].r)cnt[b[r]]--, sum -= 2 * cnt[b[r]] + 1, --r;
		mm[a[i].id] = sum;
	}
	for (int i = 0; i < m; ++i)fpn(mm[i]);
}
signed main()
{
	ll T = 1;
	//T = read();
	while (T--) {
		solve();
	}
	return 0;
}
/**
*
*　　┏┓  WQJ ┏┓+ +
*　┏┛┻━━━┛┻┓ +
*　┃　　　　　　　┃
*　┃　　　━　　　┃ ++ + + +
*  ████━████+
*  ◥██◤　◥██◤ +
*　┃　　　┻　　　┃
*　┃　　　　　　　┃ + +
*　┗━┓　　　┏━┛
*　　　┃　　　┃ + + + +Code is far away from 　
*　　　┃　　　┃ + bug with the animal protecting
*　　　┃　 　 ┗━━━┓ 神兽保佑,代码无bug　
*　　　┃ 　 　　　    ┣┓>>>
*　　  ┃ 　-　　　　 　┏┛
*　    ┗┓┓┏━┳┓┏┛ + + + +
*　　　　┃┫┫　┃┫┫
*　　　　┗┻┛　┗┻┛+ + + +
*/

 Atcoder293_补题

#include<bits/stdc++.h>
#include<unordered_map>
#include<math.h>
#include<forward_list>
#include<unordered_set>
using namespace std;
#pragma warning(disable:4996)
const double pi = acos(-1);
#define pdd pair<double,double>
#define re register
#define ll long long
#define endl '\n'
#define pii pair<ll,ll>
#define IOS std::ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fi first
#define se second
#define ull unsigned ll
#define eb emplace_back
#define pb push_back
#define power qpow
#define rad read
#define mp make_pair
#define ls (o<<1)
#define rs (o<<1|1)
#define mid (l+r>>1)
#define fpn(x) fp(x),putchar('\n')
ll mod = 1e9 + 7;
inline ll lowbit(ll x) { return x & (-x); }
inline ll  gcd(ll a, ll b) {
	return b ? gcd(b, a % b) : a;
}
inline ll exgcd(ll a, ll b, ll& x, ll& y) {
	//ax+by=1; 返回值为a,b最大公约数
	if (!b) {
		x = 1, y = 0;
		return a;
	}
	ll g = exgcd(b, a % b, y, x);
	y -= (a / b) * x;
	return g;
}
inline ll ksc(ll a, ll b, ll mod) {
	ll ans = 0;
	while (b) {
		if (b & 1) ans = (ans + a) % mod;
		a = (a + a) % mod;
		b >>= 1;
	}
	return ans;
}
ll qpow(ll a, ll b, ll modd = mod) {
	ll ans = 1;
	while (b) {
		if (b & 1)
			ans = ksc(ans, a, modd);
		a = ksc(a, a, modd);
		b >>= 1;
	}
	return ans;

}
inline ll read() {
	ll s = 0; bool w = true;
	char c = getchar();
	for (; c < '0' || c>'9'; c = getchar())if (c == '-')w = false;
	for (; c >= '0' && c <= '9'; c = getchar())s = (s << 1) + (s << 3) + (c ^ 48);
	return w ? s : -s;
}
inline void fp(ll n) {
	if (n < 0)  putchar('-'), n = -n;
	if (n > 9) fp(n / 10);
	putchar(n % 10 + 48);
}
const int N = 2e5 + 10;
struct node {
	int l, r, id, pos;
	//l,r,左右边界，id为询问编号，pos为分块所在位置
	bool operator<(const node& x) {
		if (pos == x.pos)
			if (pos & 1)return r < x.r;
			else return r > x.r;
		return pos < x.pos;
	}
}a[N];
ll q, b[N], n;
ll ans[N], sum = 0, cnt[N];
void add(int x) {
	x = b[x];
	sum -= cnt[x] * (cnt[x] - 1) * (cnt[x] - 2) / 6;
	++cnt[x];
	sum += cnt[x] * (cnt[x] - 1) * (cnt[x] - 2) / 6;
}
void del(int x) {
	x = b[x];
	//cout << cnt[x] << endl;
	sum -= cnt[x] * (cnt[x] - 1) * (cnt[x] - 2) / 6;
	--cnt[x];
	sum += cnt[x] * (cnt[x] - 1) * (cnt[x] - 2) / 6;
}
void solve() {
	n = read();
	q = read();
	ll sn = ceil(1. * n / sqrt(q));
	for (int i = 1; i <= n; ++i) {
		b[i] = read();
	}
	for (int i = 1; i <= q; ++i) {
		a[i].l = read(), a[i].r = read();
		a[i].id = i;
		//关键点
		a[i].pos = (a[i].l - 1) / sn + 1;
	}
	sort(a + 1, a + 1 + q);
	ll l = 1, r = 0;
	for (int i = 1; i <= q; ++i) {
		while (l < a[i].l)del(l++);
		while (l > a[i].l)add(--l);
		while (r < a[i].r)add(++r);
		while (r > a[i].r)del(r--);
		ans[a[i].id] = sum;
	}
	for (int i = 1; i <= q; ++i) {
		fpn(ans[i]);
	}
}
signed main()
{
	ll T = 1;
	//T = read();
	while (T--) {
		solve();
	}
	return 0;
}
/**
*
*　　┏┓  WQJ ┏┓+ +
*　┏┛┻━━━┛┻┓ +
*　┃　　　　　　　┃
*　┃　　　━　　　┃ ++ + + +
*  ████━████+
*  ◥██◤　◥██◤ +
*　┃　　　┻　　　┃
*　┃　　　　　　　┃ + +
*　┗━┓　　　┏━┛
*　　　┃　　　┃ + + + +Code is far away from 　
*　　　┃　　　┃ + bug with the animal protecting
*　　　┃　 　 ┗━━━┓ 神兽保佑,代码无bug　
*　　　┃ 　 　　　    ┣┓>>>
*　　  ┃ 　-　　　　 　┏┛
*　    ┗┓┓┏━┳┓┏┛ + + + +
*　　　　┃┫┫　┃┫┫
*　　　　┗┻┛　┗┻┛+ + + +
*/