Codeforces 483B - Friends and Presents(二分)
2014-11-01 10:28
274 查看
又是一道涨姿势的题目
这个同学解释的很好,附链接:http://www.cnblogs.com/windysai/p/4058235.html
画个韦恩图的话就很直观了
这个同学解释的很好,附链接:http://www.cnblogs.com/windysai/p/4058235.html
画个韦恩图的话就很直观了
#include <iostream> #include <cstdio> #include <cmath> using namespace std; typedef long long LL; #define maxn 1e18 LL cnt1,cnt2,x,y; bool solve(LL m){ LL f1=m/x; //能被x整除的数的个数 LL f2=m/y; //能被y整除的数的个数 LL both=m/(x*y); //既能被y也能被x整除的数的个数 LL others=m-f1-f2+both; //不能被y也不能被x整除的数的个数 LL ff1=f1-both; //只能被f1整除的数的个数 LL ff2=f2-both; //只能被f2整除的数的个数 LL gf1=(cnt1-ff2>=0 ? cnt1-ff2 : 0); //在others里能选择出可以给第一个人的数的数字个数 LL gf2=(cnt2-ff1>=0 ? cnt2-ff1 : 0); //在others里能选择出可以给第二个人的数的数字个数 return (gf1+gf2<=others); } int main(){ while(cin>>cnt1>>cnt2>>x>>y){ LL l=1,r=1e18; while(l<r){ LL m=(l+r)>>1; if(solve(m)) r=m; else l=m+1; } cout<<r<<endl; } return 0; }
相关文章推荐
- 【codeforces 483B】Friends and Presents
- codeforces B. Friends and Presents(二分+容斥)
- Codeforces-689 D Friends and Subsequences(RMQ+二分)
- codeforces 689D D. Friends and Subsequences(RMQ+二分)
- Codeforces #275 (Div. 2) B - Friends and Presents(二分大法好)
- Codeforces 689D -Friends and Subsequences (RMQ查询-> ST表+二分)
- ST表与二分 (CodeForces 689D-Friends and Subsequences)
- [Codeforces 689D] Friends and Subsequences (二分+稀疏表)
- [Codeforces #275 (Div. 2)B. Friends and Presents] 二分
- 【CODEFORCES】 B. Friends and Presents
- Codeforces Round #275 (Div. 2) B. Friends and Presents 二分+数学
- Codeforces483B——二分——Friends and Presents
- codeforces Round #361 D. Friends and Subsequences (ST表,二分)
- CF689D:Friends and Subsequences(ST表 + 二分)
- codeforces 658B-Bear and Displayed Friends
- 【CodeForces】669A - Little Artem and Presents(找规律)
- Codeforces 366D Dima and Trap Graph【二分+Dfs】
- CodeForces 669A Little Artem and Presents
- CodeForces 660F Bear and Bowling 4(斜率DP+二分)
- CodeForces 578C Weakness and Poorness 二分