Skip to main content

Round Table Meeting

                         Round Table Meeting

#include <iostream>
#include <cmath>
#include <climits>

using namespace std;


int main()
{
    int n, q;
    int x, y;
    cin >> n >> q;
    int arr[2 * n + 1];

    for(int i = 1; i <= n; i++)
    {
        cin >> arr[i];
        arr[i + n] = arr[i];

    }

    for(int k = 0; k < q; k++)
    {
        cin >> x >> y;

        int dist = INT_MAX;
        int minDist = INT_MAX;
        int posx = -1;
        int posy = -1;
       
        if(x == y)
        {
         
            minDist = 0;
           
        }
       
        else
        {
           
       
            for(int i = 1; i <= 2 * n; i++)
            {
                if(arr[i] == x)
                {
                      posx = i;
               
                }
                else if(arr[i] == y)
                {
                      posy = i;
               
                }

                if(posx != -1 && posy != -1)
                {
                   
                      dist = abs(posy - posx);
                      if(dist < minDist)
                          minDist = dist;
                }
                if(minDist <= 1)
                    break;
            }
        }
       
        if(minDist <= 1)
        {
            cout << "0\n";
        }
        else
        {
           
            cout << ceil(minDist / 2) << endl;
        }
    }

    return 0;
}

Comments

Post a Comment

Popular posts from this blog

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

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

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