1288A - Deadline - CodeForces Solution


binary search brute force math ternary search *1100

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Python Code:

from math import ceil
from math import sqrt
t = int(input())
s = {}
for i in range(t):
    n,d = map(int,input().split())
    s[i] = [n,d]

for key,value in s.items():
    n = int(value[0])
    d = int(value[1])
    if d > n: 
        flag = True
        for x in range(1,ceil(sqrt(d))): 
            total = 0
            y = ceil(float(d)/(x+1))  
            total = x + y
            if total <= n: 
                flag = True
                break
            else:
                flag = False

        if flag == True:
            print('YES')
        else:
            print('NO')
    else: 
        print('YES')
 	 	 					 		      	  	 					

C++ Code:

#include <bits/stdc++.h>

using namespace std;

int main() {
    int t,n,d;
    cin>>t;
    while(t--){
        cin>>n>>d;
        bool possible = false;
        for (int i = 0; i < n; i++)
        {
            if ((i+d)/(i+1) <= n-i) {
                possible = true;
                break;
            }
        }
        if (possible) {
            cout << "YES" << endl;
        } else {
            cout << "NO" << endl;
        }
    }
    return 0;
}


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