2018 容斥定理

2018

​​Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2006    Accepted Submission(s): 941​​

​Problem Description​

Given , find out the number of pairs of integers  where  and  is a multiple of .

​Input​

The input consists of several test cases and is terminated by end-of-file.

Each test case contains four integers .

​Output​

For each test case, print an integer which denotes the result.

## Constraint

* 

* The number of tests cases does not exceed .

​Sample Input​

1 2 1 2018 1 2018 1 2018 1 1000000000 1 1000000000

​Sample Output​

3 6051 1485883320325200

这题我怎么感觉我写过题解，无奈找不到了，重写一次吧

两个区间可以分成以下情况：

由于2018的因子只有4个，于是采用手撸容斥定理写。答案就是

左边2的的部分乘右边1009的部分

加左边1009的的部分乘右边2的部分

加左边2018部分乘右边1部分加

加左边1部分乘右边2018部分加

减去左边2018乘右边2018

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a,b,c,d;
int main()
{
    while(cin>>a>>b>>c>>d)
    {
        ll n2=b/2-(a-1)/2;
        ll n1009=b/1009-(a-1)/1009;
        ll n2018=b/2018-(a-1)/2018;
        n2-=n2018;
        n1009-=n2018;

        ll m2=d/2-(c-1)/2;
        ll m1009=d/1009-(c-1)/1009;
        ll m2018=d/2018-(c-1)/2018;
        m2-=m2018;
        m1009-=m2018;
        ll ans=n2*m1009+m2*n1009+n2018*(d-c+1)+m2018*(b-a+1)-n2018*m2018;
        printf("%lld\n",ans);
    }
}