A positive integer is magical if it is divisible by either A or B.
Return the N-th 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
Example 3:
Input: N = 5, A = 2, B = 4
Output: 10
Example 4:
Input: N = 3, A = 6, B = 4
Output: 8
Note:
1 <= N <= 10^92 <= A <= 400002 <= B <= 40000
Approach #1: Binary Serach + Brute Froce. [Time limit exceeded]
class Solution {
public:
int nthMagicalNumber(int N, int A, int B) {
long l = 1, r = N * min(A, B);
while (l <= r) {
long m = l + (r - l) / 2;
int num = 0;
for (int i = 1; i <= m; ++i) {
if (i % A == 0 || i % B == 0)
num++;
}
if (num >= N) r = m - 1;
else l = m + 1;
}
return l;
}
};
Approach #2: Binary Serach + LCM.
class Solution {
public:
int nthMagicalNumber(int N, int A, int B) {
int MOD = 1e9 + 7;
int L = A / gcd(A, B) * B;
long l = 0, r = (long) 1e15;
while (l < r) {
long m = l + (r - l) / 2;
long num = m / A + m / B - m / L; // not int type
if (num < N) l = m + 1;
else r = m;
}
return (int) (l % MOD);
}
private:
int gcd(int x, int y) {
if (x == 0) return y;
return gcd(y % x, x);
}
};
原文：https://www.cnblogs.com/ruruozhenhao/p/9940235.html