牛客练习赛52 B Galahad (树状数组)

题目链接：https://ac.nowcoder.com/acm/contest/1084/B

题意

5e5的区间，5e5个询求[l,r]区间内出现过的数的和

思路

1s时限，莫队显然会T

我们可以将询问按r排序，维护每个数最后出现的位置，并用树状数组维护前缀和即可

代码

#include<iostream>

#include<cstdio>

#include<algorithm>

#include<cmath>

#include<cstring>

#include<string>

#include<stack>

#include<queue>

#include<deque>

#include<set>

#include<vector>

#include<map>

#include<functional>

#include<unordered_map>

#define fst first

#define sc second

#define pb push_back

#define mem(a,b) memset(a,b,sizeof(a))

#define lson l,mid,root<<1

#define rson mid+1,r,root<<1|1

#define lc root<<1

#define rc root<<1|1

using namespace std;

typedef double db;

typedef long double ldb;

typedef long long ll;

typedef unsigned long long ull;

typedef pair<int,int> PI;

typedef pair<ll,ll> PLL;

const db eps = 1e-6;

const int mod = 998244353;

const int maxn = 5e5+100;

const int maxm = 2e6+100;

const int inf = 0x3f3f3f3f;

const db pi = acos(-1.0);

int n,m;

int a[maxn];

ll tree[maxn];

int lowbit(int x){return x&(-x);}

void add(int x, ll c){

    for(int i = x; i <= n; i+=lowbit(i)){

        tree[i]+=c;

    }

}

ll sum(int x){

    ll ans = 0;

    for(int i = x; i; i-=lowbit(i)){

        ans+=tree[i];

    }

    return ans;

}

struct node{

    int l, r;

    int id;

    node(){}

    node(int l, int r, int id):l(l),r(r),id(id){}

}q[maxn];

int cnt[maxn];

int id[maxn];

ll ans[maxn];

bool cmp(node a, node b){return a.r<b.r;}

int main(){

    scanf("%d %d", &n, &m);

    for(int i = 1; i <= n; i++){

        scanf("%d", &a[i]);

    }

    for(int i = 1; i <= m; i++){

        int l,r;

        scanf("%d %d", &l, &r);

        q[i]=node(l,r,i);

    }

    sort(q+1,q+1+m,cmp);

    int lst = 1;

    for(int i = 1; i <= m; i++){

        for(int j = lst; j <= q[i].r; j++){

            if(id[a[j]]){

                add(id[a[j]],-a[j]);

            }

            id[a[j]]=j;

            add(j,a[j]);

        }lst=q[i].r+1;

        ans[q[i].id]=sum(q[i].r)-sum(q[i].l-1);

    }

    for(int i = 1; i <= m; i++){

        printf("%lld\n",ans[i]);

    }

    return 0;

}

/*

3 3

1 1 1

1 3

2 3

2 2

 */