1201. Ugly Number III

Write a program to find the n-th ugly number.

Ugly numbers are positive integers which are divisible by a or b or c.

 

Example 1:

Input: n = 3, a = 2, b = 3, c = 5
Output: 4
Explanation: The ugly numbers are 2, 3, 4, 5, 6, 8, 9, 10... The 3rd is 4.

Example 2:

Input: n = 4, a = 2, b = 3, c = 4
Output: 6
Explanation: The ugly numbers are 2, 3, 4, 6, 8, 9, 12... The 4th is 6.

Example 3:

Input: n = 5, a = 2, b = 11, c = 13
Output: 10
Explanation: The ugly numbers are 2, 4, 6, 8, 10, 11, 12, 13... The 5th is 10.

Example 4:

Input: n = 1000000000, a = 2, b = 217983653, c = 336916467
Output: 1999999984

 

Constraints:

• 1 <= n, a, b, c <= 10^9
• 1 <= a * b * c <= 10^18
• It's guaranteed that the result will be in range [1, 2 * 10^9]

Discuss

思路：二分+容斥

class Solution(object):
    def nthUglyNumber(self, n, a, b, c):
        """
        :type n: int
        :type a: int
        :type b: int
        :type c: int
        :rtype: int
        """
        def gcd(a,b):
            if b==0: return a
            return gcd(b,a%b)
        a,b,c=sorted([a,b,c])[::-1]
        ab,ac,bc = a*b//gcd(a,b),a*c//gcd(a,c),b*c//gcd(b,c)
        abc = a*bc//gcd(a,bc)
#        print(ab,ac,bc,abc)
        
        def count(mid):
            res=0
            res+=(mid//a+mid//b+mid//c)
            res-=(mid//ab+mid//ac+mid//bc)
            res+=(mid//abc)
            return res
        
        
        lo,hi=1,int(1e9*2)
        while lo<hi:
            mid=(lo+hi)//2
            cnt=count(mid)
            if cnt>=n:
                hi=mid
            else:
                lo=mid+1
        return lo