HackerEarth Utkarsh and Party Row House solution

YASH PAL, 31 July 202411 February 2026

In this HackerEarth Utkarsh and Party Row House problem solution In the college where Utkarsh studies, all the students live in N Row Houses. The colony containing the Row Houses is a straight line = x axis. Ith Row House is located at x = i and Ai students live in it.

Every week the students of some Row House at x=K organize a party and invite all the students living in Row Houses from x=L to x=R.

Utkarsh wants you to handle Q events. An event can be one of the following 2 types.

K L R: Row House at x=K organizes a party and invites all students living in Row House from x=L to x=R. Tell the sum of distance traveled by all the guests.
If a guest is at x=i and party is at x=j then distance travelled by him alone is |i – j|.
K S: S new students are admitted into the college and all of them start to live in Row House at x=K

HackerEarth Utkarsh and Party Row House problem solution

HackerEarth Utkarsh and Party Row House problem solution.

#include<bits/stdc++.h>
#include<iostream>
using namespace std;
#define fre     freopen("0.in","r",stdin);freopen("0.out","w",stdout)
#define abs(x) ((x)>0?(x):-(x))
#define MOD 1000000007
#define lld signed long long int
#define pii pair<int,int>
#define scan(x) scanf("%d",&x)
#define print(x) printf("%dn",x)
#define scanll(x) scanf("%lld",&x)
#define printll(x) printf("%lldn",x)
vector<int> G[2*100000+5];
#define MAX_N 100000
lld t1[MAX_N+5];
lld t2[MAX_N+5];
lld read(lld *bit,int idx){
    lld sum = 0;
    if(idx==0)
        return 0;
    while (idx > 0){
        sum += bit[idx];
        idx -= (idx & -idx);
    }
    return sum;
}
void update(lld *bit,int idx ,lld val){
    while (idx <= MAX_N){
        bit[idx] += val;
        idx += (idx & -idx);
    }
}
lld rangeQuery(lld *bit,int i,int j){
    return read(bit,j) - read(bit,i-1);
}
int main(){
    int N,Q,L,R,K;
    lld S,x;
    cin>>N>>Q;
    for(int i=1;i<=N;++i){
        scanll(x);
        update(t1,i,x);
        update(t2,i,i*x);
    }
    while(Q--){
        int C;
        scan(C);
        if(C==1){
            scan(K);
            scan(L);
            scan(R);
            lld ans = 0;
            if(K<=L){
                ans = rangeQuery(t2,L,R) - rangeQuery(t1,L,R)*K;
            }
            else if(K<=R){
                ans = rangeQuery(t2,K,R) - rangeQuery(t1,K,R)*K
                    + rangeQuery(t1,L,K)*K - rangeQuery(t2,L,K);
            }
            else{
                ans = rangeQuery(t1,L,R)*K - rangeQuery(t2,L,R);
            }
            printll(ans);
        }
        else{
            scan(K);
            scanll(S);
            update(t1,K,S);
            update(t2,K,K*S);
        }
    }
}

Second solution

#include <iostream>
#include <cstdio>

using namespace std;

int n, q;
const int MAXN = 100100;
long long a[MAXN];
int main()
{
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; i++) {
        scanf("%lld", &a[i]);
    }
    for (int it = 1; it <= q; it++) {
        int tp;
        scanf("%d", &tp);
        if (tp == 1) {
            int k, l, r;
            scanf("%d%d%d", &k, &l, &r);
            long long res = 0;
            for (int j = l; j <= r; j++) {
                res += a[j]*abs(j - k);
            }
            printf("%lldn", res);
        }
        else {
            int wh, s;
            scanf("%d%d", &wh, &s);
            a[wh] += s;
        }

    }
    return 0;
}

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