C语言排序(6)___送数字——(二分查找)

Description

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 Input

Input

3 1 2 3

Output

5

Input

1 3 2 3

Output

4

Hint

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.

样例1：

#include <stdio.h>int x,y,n1,n2;             int ComMu(int a,int b)        //定义函数求a,b的最小公倍数{int m, n,c;  m=a;n=b;  while(b!=0)  { c=a%b;a=b; b=c;}   return m*n/a;  }int seeing (long long int top)       //定义函数判断当前数字是否满足题意{long long int a,b,c,max,book=0;max=ComMu(x,y);a=top/y-top/max;             //只能送给x的数字个数b=top/x-top/max;             //只能送给y的数字个数c=top-a-b-top/max;           //x,y都能送的数字个数if(a+c-n1>=0 )    {if(a>n1 && b+c>=n2)book=1;else if(a<=n1 && b+(c+a-n1)>=n2)book=1;}return book;          //book为1表示当前数字可以.}long long int bin(long long int l,long long int r) //二分法列举可能的数字{long long int mid;while(l<r){mid=(l+r)/2;if(seeing(mid))r=mid;elsel=mid+1;}return r;}int main(){int a=0,b=0,c=0;scanf("%d%d%d%d",&n1,&n2,&x,&y);printf("%d\n",bin(1,2000000000));return 0;}

样例2：

#include <stdio.h>int main(){int n,m,x,y;while(~scanf("%d%d%d%d",&n,&m,&x,&y))     //多组输入{int l=1,r=2e9;        while(l<r)                           //二分法枚举{              int mid=l+(r-l)/2;if(n<=(mid-mid/x)&&m<=(mid-mid/y)&&n+m<=mid-mid/(x*y)) //此判断利用数学规律,和样例1seeing函数作用等同,r=mid;else  l=mid+1;}printf("%d\n",r);}return 0;}

大概题意;

输入4个整数 n,m,x,y分别表示：至少给伙伴x的数字个数、至少给伙伴y的数字个数、伙伴x讨厌的素数x，伙伴y讨厌的素数y;

要求从1~v中分别发数字给伙伴x,伙伴y. 伙伴x不要x的倍数,伙伴y同样不要y的倍数.

求至少v为多少时,才能从1~v中发给伙伴x,伙伴y符合要求的数字分别至少为n,m个.

坑点:

在样例1方法中要注意在求最大公倍数的时候数据溢出,所以用long long 型.