深入解析：LeetCode //C - 878. Nth Magical Number

878. Nth Magical Number

A positive integer is magical if it is divisible by either a or b.

Given the three integers n, a, and b, return the nth magical number. Since the answer may be very large, return it modulo$10^9+7$.

Example 1:

Input:n = 1, a = 2, b = 3
Output: 2

Example 2:

Input:n = 4, a = 2, b = 3
Output: 6

Constraints:

• $1<=n<=10^9$
• $2<=a,b<=4\ast 10^4$

From: LeetCode
Link: 878. Nth Magical Number

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Solution:

Ideas:

• Definition: A magical number is divisible by either a or b.

• Goal: Find the n-th magical number.

• Key math:

  • Count of magical numbers ≤ x = x/a + x/b - x/lcm(a, b).

  • Use gcd to compute lcm.

• Approach:

  • Binary search the smallest number x such that count ≥ n.

  • Search range: [1, min(a, b) * n].

• Result: Return x % (10^9 + 7).

Code:

int nthMagicalNumber(int n, int a, int b) {
const long long MOD = 1000000007LL;
// gcd
long long x = a, y = b;
while (y) { long long t = x % y; x = y; y = t; }
long long g = x;
long long lcm = (long long)a / g * b;
long long lo = 1;
long long minab = a < b ? a : b;
long long hi = minab * (long long)n; // upper bound
while (lo < hi) {
long long mid = lo + (hi - lo) / 2;
long long cnt = mid / a + mid / b - mid / lcm; // numbers divisible by a or b
if (cnt >= n) hi = mid;
else lo = mid + 1;
}
return (int)(lo % MOD);
}