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== ? x : gcd(y, x%y);
 }

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

 int main(){
     long long ld = , rd = 100000000000000ll, mid;
     long long c1, c2, x, y;
     cin>>c1>>c2>>x>>y;
     while(ld <= rd){
         mid = (ld + rd)>>;
         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-;
         else ld = mid+;
     }
     cout<<rd+<<endl;
     return ;
 }