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Codeforces 483B. Friends and Presents 二分查找

2017年11月23日 ⁄ 综合 ⁄ 共 1963字 ⁄ 字号 评论关闭

二分查找:

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 andcnt2 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 cnt1cnt2xy (1 ≤ cnt1, cnt2 < 109cnt1 + cnt2 ≤ 1092 ≤ x < y ≤ 3·104) —
the numbers that are described in the statement. It is guaranteed that numbers xy 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 135 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.

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

using namespace std;

typedef long long int LL;

LL c1,c2,x,y;

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

bool ck(LL n)
{
    if(n-n/x<c1||n-n/y<c2) return false;
    LL all=n-n/(x/gcd(x,y)*y);
    if(all>=c1+c2) return true;
    return false;
}

LL bin()
{
    LL low=1,high=120000000000LL,mid,ans;
    while(low<=high)
    {
        mid=(low+high)/2;
        if(ck(mid))
        {
            ans=mid; high=mid-1;
        }
        else low=mid+1;
    }
    return ans;
}

int main()
{
    cin>>c1>>c2>>x>>y;
    cout<<bin()<<endl;
    return 0;
}

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