Codeforces 750A - New Year and Hurry

Problem Link - https://codeforces.com/contest/750/problem/A

Editorial

Do you see what is produced by the following piece of code?

int total = 0;
for(int i = 1; i <= n; ++i) {
	total += 5 * i;
	printf("%d\n", total);
}

We iterate over problems (a variable i denotes the index of problem) and in a variable total we store the total time needed to solve them. The code above would print numbers 5, 15, 30, 50, ... — the i-th of these numbers is the number of minutes the hero would spend to solve easiest i problems.

Inside the loop you should also check if there is enough time to make it to the party, i.e. check if total + k <= 240.

Solution

        

    #include<bits/stdc++.h>

    using namespace std;

    int main()

    {  

    int n,k; int count=0; int sum=0;

    cin>>n>>k;

    for(int i=1; i<=n; i++)

    {

        sum += 5*i;  

        if(sum>240-k) break;

        count +=1;

    }

    cout<<count<<endl;

    return 0;

    }

Codeforces 556A ( Case of Zeros and Ones )

Codeforces 556A

Case of Zeros and Ones

Problem Link:
http://codeforces.com/problemset/problem/556/A

Codeforces 791A ( Bear and Big Brother )

Codeforces 791A 

( Bear and Big Brother )

Problem Link:
http://codeforces.com/problemset/problem/791/A

Solution : 

#include<bits/stdc++.h>

using namespace std;

int main()

{

    int a,b,i,j,cnt;

    cin>>a>>b;

    cnt=0;

    for(i=1;; i++)

    {

            a=a*3;

            b=b*2;

            cnt++;

        if(a>b)

        {

            break;

        }

    }

    cout<<cnt<<endl;

    return 0;

}

Codeforces Problem 762A (K-th-divisor)

Codeforces Problem 762A 

(K-th-divisor)

Problem link : 
http://codeforces.com/problemset/problem/762/A

You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.

Divisor of n is any such natural number, that n can be divided by it without remainder.

Input

The first line contains two integers n and k (1 ≤ n ≤ 10¹⁵, 1 ≤ k ≤ 10⁹).

Output

Codeforces 749A (Bachgold Problem)

Codeforces 749A (Bachgold Problem)

time limit per test-

1 second

memory limit per test-

256 megabytes

input-

standard input

output-

standard output

Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.

Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.