Skip to main content

IPC Trainers : IPCTRAIN

Problem Link

#include <iostream>
#include <vector>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;

const int N = 1e5;

struct node
{
    int trainer;
    int day;
    int sadness;

    node()
    {

    }
    node(int t, int d, int s)
    {
        trainer = t;
        day = d;
        sadness = s;
    }
};

bool comparator(node a, node b)
{
    return a.sadness > b.sadness;
}
int main()
{
    long long result;
    int test, n, D, d, t, s;
    int freq[N + 1], sad[N + 1];
    vector<node> v;
    node temp;

    cin >> test;

    while(test--)
    {
        cin >> n >> D;
        memset(freq, 0, sizeof freq);
        memset(sad, 0, sizeof sad);
        result = 0;

        v.clear();
        set<int> daySet;

        for (int i = 0; i < D; i++) {
daySet.insert(i);
}

        for(int i = 0; i < n; i++)
        {
            cin >> d >> t >> s;
            freq[i] = t;
            sad[i] = s;
            v.push_back(node(i, d - 1, s));
        }

        sort(v.begin(), v.end(), comparator);

        for(int j = 0; j < n; j++)
        {
            temp = v[j];

            while(freq[temp.trainer] > 0)
            {
                auto iter = daySet.lower_bound(temp.day);
                if(iter == daySet.end())
                    break;
                else
                {
                    freq[temp.trainer]--;
                    daySet.erase(iter);
                }
            }
        }

        for(int i = 0; i < n; i++)
        {
            result += (long long)sad[i] * (long long) freq[i];
        }

        cout << result << endl;
    }


    return 0;
}

Comments

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++)  

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; } }

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