Skip to main content

Architect's Dilemma..!

                                     Architect's Dilemma..!

                                                                    Binary Search


#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;

ll n, w, a[100005], pre[100005], maxx[100005];

bool f(int x)
{
    for(int i = 0, j = x; j <= n; j++, i++)
    {
        if((x * maxx[j] - (pre[j] - pre[i])) <= w)
            return 1;
    }
    return 0;
}

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

    sort(a + 1, a + n + 1);
    for(int i = 1; i <= n; i++)
    {
        pre[i] = pre[i - 1] + a[i];
        maxx[i] = max(maxx[i - 1], a[i]);
    }

    int l = 1, r = n, ans;

    while(l <= r)
    {
        int mid = (l + r) / 2;
        if(f(mid))
        {
            ans = mid;
            l = mid + 1;
        }
        else
            r = mid - 1;
    }

    cout << ans << 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 ...

ANARC09A

ANARC09A - Seinfeld #include <iostream> #include <stack> using namespace std; stack<char> st; int main() {     string s;     int c, k;     k = 1;     while(true)     {         cin >> s;         c = 0;         if(s[0] == '-')             break;         for(int i = 0; i < s.size(); i++)         {             st.push(s[i]);             //stack contains } and { then we can safely remove them from stack             if(st.top() == '}' && st.size() != 1)             {                 char a = st.top();               ...

CRDS

CRDS - Cards #include <iostream> using namespace std; long int m = 1000007; int main() {     int t;     long long int n;     long long int result;     cin >> t;     while(t--)     {         cin >> n;         result = 0;         result += 3 * (n * (n + 1) / 2) - n;         result = result % m;         cout << result << endl;     }     return 0; }