題目描述
Given an integer num, find the closest two integers in absolute difference whose product equals num + 1 or num + 2.
Return the two integers in any order.
Example 1:
Input: num = 8 Output: [3,3] Explanation: For num + 1 = 9, the closest divisors are 3 & 3, for num + 2 = 10, the closest divisors are 2 & 5, hence 3 & 3 is chosen.
Example 2:
Input: num = 123 Output: [5,25]
Example 3:
Input: num = 999 Output: [40,25]
Constraints:
1 <= num <= 10^9
解析
對於數num，若i*j==num，假設i<=j，則必有i<=sqrt(num)、j>=sqrt(num)。
class Solution {
public:
vector<int> closestDivisors(int num) {
vector<int> ans(2);
int sqr=sqrt(num+2);
for(int i=sqr;i>=1;i--){
if((num+1)%i==0){
ans[0]=i;
ans[1]=(num+1)/i;
break;
}
else if((num+2)%i==0){
ans[0]=i;
ans[1]=(num+2)/i;
break;
}
}
return ans;
}
};