【CODEFORCES】 B. Friends and Presents

解决如何在限制条件下为两位朋友分配互不相同的整数礼物的问题。利用二分查找与容斥原理,确保分配的礼物数符合特定条件且不超过指定范围。
B. Friends and Presents
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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 cnt1cnt2xy (1 ≤ cnt1, cnt2 < 109cnt1 + cnt2 ≤ 1092 ≤ x < y ≤ 3·104) — the numbers that are described in the statement. It is guaranteed that numbers xy are prime.

Output

Print a single integer — the answer to the problem.

Sample test(s)
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 135 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:这一题是2分加容斥原理,二分整个区间,这里取的上界是2e9。这样一个人不收x的倍数,另一个人不收y的倍数,对于一个解n来说,n-n/x就是第一个人最多能接受的个数。第二个人同理。又因为所有的数中可能有些数即是x的倍数又是y的倍数,所以要用到容斥原理,算出n中总共有多少数是两个人都能接受的。算出这些之后与cnt比较就可以二分了。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

long long cnt1,cnt2,x,y,t;

long long gcd(long long x,long long y)
{
	if (y==0) return x;
	else return gcd(y,x%y);
}

int main()
{
    scanf("%I64d%I64d%I64d%I64d",&cnt1,&cnt2,&x,&y);
    long long l=1,r=2000000000,k=x*y/gcd(x,y);
    while (l<r)
    {
        if (l==r)
        {
            printf("%I64d",r);
            return 0;
        }
        t=(l+r)/2;
        if (cnt1<=t-t/x && cnt2<=t-t/y && t>=cnt1+cnt2+t/k ) r=t;
        else l=t+1;
    }
    printf("%I64d",r);
    return 0;
}


### Codeforces Div.2 比赛难度介绍 Codeforces Div.2 比赛主要面向的是具有基础编程技能到中级水平的选手。这类比赛通常吸引了大量来自全球不同背景的参赛者,包括大学生、高中生以及一些专业人士。 #### 参加资格 为了参加 Div.2 比赛,选手的评级应不超过 2099 分[^1]。这意味着该级别的竞赛适合那些已经掌握了一定算法知识并能熟练运用至少一种编程语言的人群参与挑战。 #### 题目设置 每场 Div.2 比赛一般会提供五至七道题目,在某些特殊情况下可能会更多或更少。这些题目按照预计解决难度递增排列: - **简单题(A, B 类型)**: 主要测试基本的数据结构操作和常见算法的应用能力;例如数组处理、字符串匹配等。 - **中等偏难题(C, D 类型)**: 开始涉及较为复杂的逻辑推理能力和特定领域内的高级技巧;比如图论中的最短路径计算或是动态规划入门应用实例。 - **高难度题(E及以上类型)**: 对于这些问题,则更加侧重考察深入理解复杂概念的能力,并能够灵活组合多种方法来解决问题;这往往需要较强的创造力与丰富的实践经验支持。 对于新手来说,建议先专注于理解和练习前几类较容易的问题,随着经验积累和技术提升再逐步尝试更高层次的任务。 ```cpp // 示例代码展示如何判断一个数是否为偶数 #include <iostream> using namespace std; bool is_even(int num){ return num % 2 == 0; } int main(){ int number = 4; // 测试数据 if(is_even(number)){ cout << "The given number is even."; }else{ cout << "The given number is odd."; } } ```
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