CodeForce l-Friends and Presents

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.

Example
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.

题解：题意即是给1号小朋友分cnt1个数字（不能分x的倍数），2号小朋友分cnt2个数字（不能是y的倍数，由于要求v使得v最小满足题意，可采用二分法去做

AC1

#include<bits/stdc++.h>typedef long long LL;using namespace std;int main(){    int cnt1,cnt2,x,y;    scanf("%d%d%d%d",&cnt1,&cnt2,&x,&y);    LL l=0,r=1e15;    while(l<=r)    {        LL mid=(l+r)>>1;        LL rx=mid/x,ry=mid/y,re=mid/(x*y);        if(mid-re>=cnt1+cnt2&&mid-rx>=cnt1&&mid-ry>=cnt2)  r=mid-1;        else l=mid+1;    }    printf("%I64d\n",l);    return 0;}rx 代表可被x整除的个数，ry: 可被y整除的数个数，re:既可以被x,y整除的个数

AC2

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>using namespace std;typedef __int64 LL;LL cnt1, cnt2, x, y;bool check(LL v){    LL f1 = v / x;    LL f2 = v / y;    LL both = v / (x*y);    LL others = v - f1 - f2 + both;    LL ff1 = f1 - both;  // second    LL ff2 = f2 - both;  // first    LL gf1 = (cnt1 - ff2 >= 0 ? cnt1 - ff2 : 0);   // 注意这个相减有可能为负数，所以要判断下    LL gf2 = (cnt2 - ff1 >= 0 ? cnt2 - ff1 : 0);    return (gf1 + gf2 <= others);}int main(){    while (scanf("%I64d%I64d%I64d%I64d", &cnt1, &cnt2, &x, &y) != EOF)    {        LL l = 1, r = 1e18;        while (l < r)        {            LL m = (l+r) >> 1;            if (check(m))                r = m;            else                l = m + 1;        }        printf("%I64d\n", r);    }    return 0;}

思想一样，只是思路更加清晰点：
1. f1: 在 1-v 中可以被x 整除的个数
2. f2: 在 1-v 中可以被y 整除的个数
3. both: 既可以被x,也可以被整除
4. other ：都不可以整除的数
5. ff1: 能分配第二个人但不是从others 中选出来的
6. ff2: 能分配给第一个人但不是从others 中选择出来的
7. cnt1=ff2+others
8. cnt2=ff1+others