codeforces B. Friends and Presents(二分+容斥)

题意：从1....v这些数中找到c1个数不能被x整除，c2个数不能被y整除！
并且这c1个数和这c2个数没有相同的！给定c1, c2, x, y， 求最小的v的值！

思路： 二分+容斥，二分找到v的值，那么s1 = v/x是能被x整除的个数
s2 = v/y是能被y整除数的个数，s3 = v/lcm(x, y)是能被x,y的最小公倍数
整除的个数！
那么 v-s1>=c1 && v-s2>=c2 && v-s3>=c1+c2就是二分的条件！

#include<iostream> 
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

int gcd(int x, int y){
    return y==0 ? x : gcd(y, x%y);
}

int lcm(int x, int y){
    return x*y/gcd(x, y);
}

int main(){
    long long ld = 1, rd = 100000000000000ll, mid;
    long long c1, c2, x, y;
    cin>>c1>>c2>>x>>y;
    while(ld <= rd){
        mid = (ld + rd)>>1;
        long long s1 = mid/x, s2 = mid/y, s3 = mid/lcm(x, y);
        if(mid-s1 >= c1 && mid-s2 >= c2 && mid-s3 >= c1+c2) rd = mid-1;
        else ld = mid+1;
    }    
    cout<<rd+1<<endl;
    return 0;
}