B. Friends and Presents（Codeforces Round #275(div2)_html/css_WEB-ITnose

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 cnt1, cnt2, x, y (1?≤?cnt1,?cnt2?109; cnt1?+?cnt2?≤?109; 2?≤?x?

Output

Print a single integer ? the answer to the problem.

Sample test(s)

input

3 1 2 3

output

input

1 3 2 3

output

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.

二分渣渣把二分又写跪了，总是分不清l与r的关系o(?□?)o，我竟然l和r都判断了一下。这题竟然l和r都能过。

这题就是枚举一下中间结果，对a周期为x-1，b周期为y-1，只有当i为y的倍数时，只能让a取，当i为x的倍数时，只

能让b取，算一下x，y的倍数时两个都不能取得，a的总数量减去只能a取的，b的总数量减去只能b取的 ，剩下的要

取的和小于等于两个都能取得，这个值就是有效值。

代码：

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;long long cnt1,cnt2,x,y;long long gcd(long long a,long long b){    return b==0?a:gcd(b,a%b);}bool judge(long long v){    long long t1,t2,t3;    t1=v/x;    t2=v/y;    t3=v/(x*y/gcd(x,y));    long long temp1,temp2,temp3;    temp1=max((long long)0,cnt1-t2+t3);//t2-t3是只能a取的，cnt1-只能a取的，就是剩下a没取的    temp2=max((long long)0,cnt2-t1+t3);//t1-t3是只能b取的，cnt2-只能b取的，就是剩下b没取的    temp3=max((long long)0,v-t1-t2+t3);//x，y都能取的    if(temp3>=temp1+temp2)//a，b都能取的大于要大于等于a，b没取的和    return true;    else    return false;}int main(){    scanf("%I64d%I64d%I64d%I64d",&cnt1,&cnt2,&x,&y);    long long l=1,r=2000000000;    long long u=0;    while(l<r int m="(l+r)">>1;        if(judge(m))        {            u=m;            r=m;        }        else        {            l=m+1;        }    }   // long long ans;   // if(judge(r))   // ans=r; //   else  //  ans=l;    printf("%I64d\n",u);    return 0;}</r></algorithm></cstdio></iostream>