Codeforces483B. Friends and Presents（二分+容斥原理）

题目链接：传送门

题目：

B. Friends and Presents
time limit per test
 second
memory limit per test
 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 , , ..., v. Of course you may choose not to present some numbers at all.

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

The only line contains four positive integers cnt1, cnt2, x, y ( ≤ cnt1, cnt2 < ; cnt1 + cnt2 ≤ ;  ≤ x < y ≤ ·) — 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

Output
Copy

Input
Copy

Output
Copy

Note

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

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

题目大意：

　　已知素数x，y，要求从1开始分别分配cnt1，cnt2个数给x，y，且分配给x的数不能是x的倍数，分配给y的数不能是y的倍数。求所有分掉的数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）：

　　那么1-mid之间所有x的倍数不能分给x，那么优先分给y；

　　同理：y的倍数都先分给x。当然lcm(x, y) = xy的倍数不能分，要减去这部分（容斥）。

　　然后比较mid的没分配的部分，和cnt1，cnt2没分到的部分。

　　这样可以用O(n)的时间验证答案，且答案是单调的，故用二分搞。

代码：

#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));
    ll resy = max(cnt2 - mul_of_x + mul_of_xy, (ll));
    return resx + resy <= tmp;
}

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