Skip to main content

KOPC12A

KOPC12A - K12 - Building Construction


#include <iostream>
#include <cmath>

#define REP(i, n) for(int i = 0; i < n; i++)

using namespace std;

const int N = 10005;

int height[N];
int cost[N];

int n;

//finds the total cost for height h
long long int findCost(int h)
{
    long long c = 0; //cost

    REP(i, n)
    {
        c += abs(h - height[i]) * cost[i];

    }

    return c;

}


int ternary_search(int l, int h)
{

    while(l <= h)
    {

        if(l == h)
            break;

        int mid1 = l + (h - l) / 3;
        int mid2 = h - (h - l) / 3;


        if(findCost(mid1) > findCost(mid2))
            l = mid1 + 1;
        else
            h = mid2 - 1;

    }

    return l;
}

int main()
{
    int t, maxHeight;

    cin >> t;

    while(t--)
    {
        cin >> n;

        maxHeight = 0;

        REP(i, n)
        {
            cin >> height[i];

            if(height[i] > maxHeight)
                maxHeight = height[i];
        }

        REP(i, n)
        {
            cin >> cost[i];
        }

        int h = ternary_search(0, maxHeight);

        cout << findCost(h) << endl;
    }
    return 0;
}

Comments

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 ...

Yet Another Cute Girl - PRETNUM

Problem Link #include <iostream> #include <cstring> #include <cmath> #include <vector> using namespace std; #define LL long long const int N = 1000010; bool prime[N + 1]; vector<int> v; void sieve() { for(int i = 2; i <= N; i++) { prime[i] = true; } for(int i = 2; i * i <= N; i++) { if(prime[i]) { for(int j = i * i; j <= N; j += i) { prime[j] = false; } } } v.push_back(2); for(int i = 3; i <= N; i += 2) { if(prime[i]) { v.push_back(i); } } } LL getNumDiv(LL num) {     /* if prime factor of n is p1^k1 p2 ^ k2 then prime factor of n^2 would be         (p1^k1 p2 ^ k2)^2 = p1^(2k1) p2^(2k2)         so number of divisors of n^2 would be (2 * k1 + 1) * (2 * k2 + 1) ...         if n is a prime number number then number of divisors of n^2 would be 3.     */ LL res = 1; for(int i = 0;...