題目連結：傳送門

題目：

B. Friends and Presents
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

You have two friends. You want to present each of them several positive integers. You want to present cnt1 numbers to the first friend and cnt2 numbers to the second friend. Moreover, you want all presented numbers to be distinct, that also means that no number should be presented to both friends.

In addition, the first friend does not like the numbers that are divisible without remainder by prime number x. The second one does not like the numbers that are divisible without remainder by prime number y. Of course, you
 
're not going to present your friends numbers they don't like.

Your task is to find such minimum number v, that you can form presents using numbers from a set 1, 2, ..., v. Of course you may choose not to present some numbers at all.

A positive integer number greater than 1 is called prime if it has no positive divisors other than 1
 
 and itself.
Input

The only line contains four positive integers cnt1, cnt2, x, y (1 ≤ cnt1, cnt2 < 109; cnt1 + cnt2 ≤ 109; 2 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers x, y are prime.
Output

Print a single integer — the answer to the problem.
Examples
Input
Copy

 
3 1 2 3

Output
Copy

5

Input
Copy

1 3 2 3

Output
Copy

4

Note

In the first sample you give the set of numbers {1, 3, 5} to the first friend and the set of numbers {2} to the second friend. Note that if you give set {1, 3, 5} to the first friend, then we cannot give any of the numbers 1, 3, 5 to the second friend.

In the second sample you give the set of numbers {3} to the first friend, and the set of numbers {1, 2, 4} to the second friend. Thus, the answer to the problem is 4.

題目大意：

　　已知素數x，y，要求從1開始分別分配cnt1，cnt2個數給x，y，且分配給x的數不能是x的倍數，分配給y的數不能是y的倍數。求所有分掉的數中的最大值的最小值。

　　1 ≤ cnt₁, cnt₂ < 10⁹; cnt₁ + cnt₂ ≤ 10⁹; 2 ≤ x < y ≤ 3·10⁴

思路：

　　如果已知答案mid（滑稽）：

　　那麼1-mid之間所有x的倍數不能分給x，那麼優先分給y；

　　同理：y的倍數都先分給x。當然lcm(x, y) = xy的倍數不能分，要減去這部分（容斥）。

　　然後比較mid的沒分配的部分，和cnt1，cnt2沒分到的部分。

　　這樣可以用O(n)的時間驗證答案，且答案是單調的，故用二分搞。

程式碼：

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

ll cnt1, cnt2, x, y;

bool judge(ll mid) {
    ll mul_of_x = mid/x;
    ll mul_of_y = mid/y;
    ll mul_of_xy = mid/x/y;
    ll tmp = mid - mul_of_x - mul_of_y + mul_of_xy;
    ll resx = max(cnt1 - mul_of_y + mul_of_xy, (ll)0);
    ll resy = max(cnt2 - mul_of_x + mul_of_xy, (ll)0);
    return resx + resy <= tmp;
}

int main()
{
    cin >> cnt1 >> cnt2 >> x >> y;
    ll l = 0, r = 1e18;
    ll ans = r;
    while (l <= r) {
        ll mid = (l+r) >> 1;
        if (judge(mid)) {
            ans = min(ans, mid);
            r = mid-1;
        }
        else
            l = mid+1;
    }
    cout << ans << endl;
    return 0;
}