【洛谷】 P2709 小B的询问 莫队算法

题意：若干询问，每次询问输出区间内每个数出现次数的平方和

思路：

既然是考莫队算法，我们就想add和del函数怎么写。考虑对每个位置i，它对答案的贡献就是这个位置产生的次数贡献，cnt[a[i]]++，然后看看应该增多少。
若原来的cnt[a[i]]为m-1，目前为m，那增量为m² - (m-1)² = 2m - 1。所以add函数就加上2*m-1即可。del则反过来。

AC代码：

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include <queue>
#include<sstream>
#include <stack>
#include <set>
#include <bitset>
#include<vector>
#define FAST ios::sync_with_stdio(false)
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<=n;++i)
#define per(i,n,a) for(int i=n;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> PII;
const int maxn = 1e5+200;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1.0);
const int mod = 1e9+7;
inline int lowbit(int x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}
inline ll inv(ll x,ll p){return qpow(x,p-2,p);}
inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }
int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

ll n,m,k;
ll a[maxn];
ll cnt[maxn];
ll pos[maxn];
ll Ans[maxn];
ll sum = 0;

typedef struct Query
{
    ll l;
    ll r;
    ll id;
}P;
P q[maxn];

bool cmp(P a, P b)
{
    if(pos[a.l] != pos[b.l]) return pos[a.l] < pos[b.l];
    if(pos[a.l]&1) return pos[a.r] < pos[b.r];
    return pos[a.r] > pos[b.r];
}

inline void add(ll x)
{
    cnt[a[x]] ++;
    sum +=  cnt[a[x]]*2 - 1;
}

inline void del(ll x)
{
    if(cnt[a[x]])
    sum -= cnt[a[x]] * 2 - 1,cnt[a[x]] --;
}

int main()
{
    n = read(), m = read(), k = read();
    ll block = pow(n,2/3.0);
    rep(i,1,n) a[i] = read(), pos[i] = (i-1)/block + 1;
    rep(i,1,m) q[i].l = read(), q[i].r = read(), q[i].id = i;
    sort(q+1,q+1+m,cmp);
    ll L = 0, R = 0;
    rep(i,1,m)
    {
        while(q[i].l < L) add(--L);
        while(q[i].r > R) add(++R);
        while(q[i].l > L) del(L++);
        while(q[i].r < R) del(R--);
        Ans[q[i].id] = sum;
    }
    rep(i,1,m) printf("%lld\n",Ans[i]);

    return 0;
}