Skip to main content

KQUERY

KQUERY - K-query


#include <iostream>
#include <algorithm>

#define pf(i) printf("%ld\n", i)
#define sf(i) scanf("%d", &i)
#define sl(i) scanf("%ld", &i)

using namespace std;

int n, q;
long int l, r, k;

struct node
{
    int l, r, p;
    int v;
};


node makenode(int v,int l, int r, int p)
{
    node temp;
    temp.v = v; //value
    temp.l = l; //left limit
    temp.r = r; //right limit
    temp.p = p; //position

    return temp;
}

bool comp(node a, node b)
{
    if(a.v == b.v)
    {
        return a.l > b.l;
    }

    return a.v > b.v;
}

void update(long int seg[], int l, int r, int index, int pos)
{
    if(l == r)
    {
        seg[index]++;
    }
    else
    {
        int mid = (l + r) / 2;

        if(pos <= mid)
        {
            update(seg, l, mid, 2 * index + 1, pos);
        }
        else
        {
            update(seg, mid + 1, r, 2 * index + 2, pos);
        }

        seg[index] = seg[2 * index + 1] + seg[2 * index + 2];
    }
}

long int query(long int seg[], int l, int r, int qs, int qe, int index)
{
    if(qs > r || qe < l)
        return 0;

    if(qs <= l && qe >= r)
        return seg[index];

    else
    {
        int mid = (l + r) / 2;

        long int p1 = query(seg, l, mid, qs, qe, 2 * index + 1);
        long int p2 = query(seg, mid + 1, r, qs, qe, 2 * index + 2);

        return p1 + p2;
    }
}

node arr[250000];
long int ans[250000];

int main()
{
    sf(n);

    long int seg[120000] = {0}; // max size 4 * n = 120000

    int index  = 0;
    for(int i = 0; i < n; i++)
    {
        sl(k);
        arr[index++] = makenode(k, 0, 0, i);
    }

    sf(q);

    for(int i = 0; i < q; i++)
    {
        sl(l),sl(r), sl(k);
        arr[index++] = makenode(k, l, r, i);
    }

    sort(arr, arr + index, comp);

    for(int i = 0; i < index; i++)
    {
        if(arr[i].l == 0) // means it is an array element
        {
            update(seg, 0, n - 1, 0, arr[i].p);
        }
        else // it is a query
        {
            ans[arr[i].p] = query(seg, 0, n - 1, arr[i].l - 1, arr[i].r - 1, 0);
        }
    }

    for(int i = 0; i < q; i++)
    {
        pf(ans[i]);
    }

    return 0;
}

Comments

Post a Comment

Popular posts from this blog

GCJ101BB - Picking Up Chicks

Problem Link /* explanation     lets solve the problem only for 2 chicken.     s[i] = speed of chicken i     pos[i] = position of chicken i     if s[i] > s[i - 1] then no problem, just check whether both can reach b within time or not.     if s[i] < s[i - 1] then there is a chance that i can slow down i - 1.     lets say s[i] = 1 m/sec and s[i - 1] = 2 m/sec and time limit is T and point to reach is B.     for s[i] pos[i] can be at max B - T. if pos is greater than B-T it can not reach within Tsec.     and for s[i - 1] pos[i - 1] can be at max (B-T)*2. if pos[i - 1] > (B-T)*2 it can not reach within Tsec.     at T sec i will be at B and i - 1 will also be at B. at T - 1 i will be at B-T-1 and i-1 will be at B-T-2 and so on. as we can see i -1 will always be behind i. so there will not be any collision.     if i is pos[i] < B-T then i can reach B before T sec and it will not cause any problem .     problem may occur if pos[i - 1]< (B -T)*2 as it can me
Post a Comment
Read more

War of XORs- XORIER

Problem Link #include <iostream> using namespace std; int main() { int t, n, odd, even; cin >> t; while(t--) { cin >> n; int i,arr[n],freq[1100001]={0}; long res = 0; odd = even = 0; for(int i = 0; i < n; i++) { cin >> arr[i]; freq[arr[i]]++; } for(int i = 0; i < n; i++) { if(arr[i] & 1) { odd++; } else { even++; } } for(int i = 0; i < n; i++) { if(arr[i] % 2) { res += odd; } else { res += even; } res -= freq[arr[i] ^ 2]; res -= freq[arr[i]]; } cout << res / 2 << endl; } }
Post a Comment
Read more

MAIN72

MAIN72 - Subset sum #include <iostream> #include <cstring> using namespace std; bool dp[100001][1001]; int arr[1001]; int main() {     int t, n;     long long int sum;     cin >> t;     while(t--)     {         cin >> n;         memset(dp, false, sizeof(dp));         sum = 0;         for(int i = 0; i < n; i++)         {             cin >> arr[i];             sum += arr[i];         }         for(int i = 0; i < n; i++)             dp[0][i] = true; // 0 sum         for(int i = 1; i < n; i++)             dp[i][0] = false; // sum is i but 0 element         for(long int i = 1; i <= sum; i++)         {             for(int j = 1; j <= n; j ++)             {                 dp[i][j] = dp[i][j - 1];                 if(i >= arr[j - 1])                     dp[i][j] = dp[i][j] || dp[i - arr[j - 1]][j - 1];             }         }         long int result = 0;         for(int i = 1; i <= sum; i++)  
Post a Comment
Read more