HDU 6286 容斥

2018

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 504    Accepted Submission(s): 262

Problem Description

 

Given  a,b,c,d, find out the number of pairs of integers (x,y) where a≤x≤b,c≤y≤d and x⋅y is a multiple of 2018.

 

Input

 

The input consists of several test cases and is terminated by end-of-file. Each test case contains four integers  a,b,c,d.

 

Output

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

*  1≤a≤b≤1e9,1≤c≤d≤1e9
* The number of tests cases does not exceed 1e4.

 

Sample Input

 

1 2 1 2018

1 2018 1 2018

1 1000000000 1 1000000000

 

Sample Output

3

6051

1485883320325200

 

Source

CCPC2018-湖南全国邀请赛-重现赛（感谢湘潭大学）

 

解析：

讨论左边区间的可能:

1.2018的倍数

2.1009的倍数但不是2018的倍数

3.2的倍数但不是2018的倍数

4. 1的倍数但不是1009与2的倍数

右边区间分别讨论：

1.1的倍数

2.2的倍数

3.1009的倍数

4.2018的倍数

代码：

#include"bits/stdc++.h"
#define ll long long
#define cl(x) scanf("%lld",&x)
#define pl(x) printf("%lld\n",x)
using namespace std;
int main()
{
    ll a,b,c,d;
    while(scanf("%lld%lld%lld%lld",&a,&b,&c,&d)==4)
    {
        ll ans=0;
        ll a1,a2,a3;
        a1=(b/2018)-((a-1)/2018);
        a1*=d-c+1;

        a2=b/1009-(a-1)/1009-(b/2018-(a-1)/2018);
        a2*=d/2-(c-1)/2;

        a3=b/2-(a-1)/2-(b/2018-(a-1)/2018);
        a3*=d/1009-(c-1)/1009;

        ans+=(b-a+1-(b/2-(a-1)/2)-(b/1009-(a-1)/1009)+(b/2018-(a-1)/2018))*(d/2018-(c-1)/2018);
        ans+=a1+a2+a3;

        pl(ans);
    }
}