【CODEFORCES】 B. Friends and Presents

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.
Sample test(s)
input

3 1 2 3

output

5

input

1 3 2 3

output

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.
문제풀이: 이 문제는 2점 가용척 원리이고 2점 전체 구간이다. 여기서 얻은 상계는 2e9이다.이렇게 한 사람은 x의 배수를 받지 않고 다른 사람은 y의 배수를 받지 않는다. n을 풀 때 n-n/x는 첫 번째 사람이 가장 많이 받아들일 수 있는 개수다.두 번째 사람은 같은 이치이다.또한 모든 수 중 x의 배수이자 y의 배수가 있을 수 있기 때문에 용척 원리를 써서 n에 모두 몇 개의 수가 있는지 계산해 내는 것은 두 사람이 모두 받아들일 수 있는 것이다.이것들을 계산한 후 cnt와 비교하면 2점이 된다.

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

long long cnt1,cnt2,x,y,t;

long long gcd(long long x,long long y)
{
	if (y==0) return x;
	else return gcd(y,x%y);
}

int main()
{
    scanf("%I64d%I64d%I64d%I64d",&cnt1,&cnt2,&x,&y);
    long long l=1,r=2000000000,k=x*y/gcd(x,y);
    while (l<r)
    {
        if (l==r)
        {
            printf("%I64d",r);
            return 0;
        }
        t=(l+r)/2;
        if (cnt1<=t-t/x && cnt2<=t-t/y && t>=cnt1+cnt2+t/k ) r=t;
        else l=t+1;
    }
    printf("%I64d",r);
    return 0;
}