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!

 1 #include<iostream> 
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<algorithm>
 5 using namespace std;
 6 
 7 int gcd(int x, int y){
 8     return y==0 ? x : gcd(y, x%y);
 9 }
10 
11 int lcm(int x, int y){
12     return x*y/gcd(x, y);
13 }
14 
15 int main(){
16     long long ld = 1, rd = 100000000000000ll, mid;
17     long long c1, c2, x, y;
18     cin>>c1>>c2>>x>>y;
19     while(ld <= rd){
20         mid = (ld + rd)>>1;
21         long long s1 = mid/x, s2 = mid/y, s3 = mid/lcm(x, y);
22         if(mid-s1 >= c1 && mid-s2 >= c2 && mid-s3 >= c1+c2) rd = mid-1;
23         else ld = mid+1;
24     }    
25     cout<<rd+1<<endl;
26     return 0;
27 } 
View Code

 

posted @ 2014-10-25 12:58  胡峻峥  阅读(329)  评论(0编辑  收藏