2018

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 500    Accepted Submission(s): 258

 

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.

## Constraint

* 1≤a≤b≤109,1≤c≤d≤109
* The number of tests cases does not exceed 104.

Sample Input

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

Sample Output

3 6051 1485883320325200

Source

#include<stdio.h>
int main()
{
	long long a,b,c,d;
	long long a1,b1,a2,b2,a3,b3;
	long long sum;
	while(~scanf("%lld%lld%lld%lld",&a,&b,&c,&d))
	{
		sum=0;
		a1=b/2018-a/2018;//區間一中2018的倍數的個數
		if(a%2018==0)//注意！！
		{
			a1++;
		}
		b1=d/2018-c/2018;//區間二中2018的倍數的個數
		if(c%2018==0)//注意！！
		{
			b1++;
		}
		sum=a1*(d-c+1)+b1*(b-a+1)-a1*b1;//注意！！a1*b1為區間一和區間二中重複
		a2=b/1009-a/1009;//區間一中1019的倍數的個數
		if(a%1009==0)//注意！！
		{
			a2++;
		}
		b2=d/1009-c/1009;//區間二中1019的倍數的個數
		if(c%1009==0)//注意！！
		{
			b2++;
		}
		a2=a2-a1;//區間一中1019的倍數的個數-區間一中2018的倍數的個數(即1019的偶數倍數)   可得1019奇數倍數
		b2=b2-b1;//區間二中1019的倍數的個數-區間二中2018的倍數的個數(即1019的偶數倍數)   可得1019奇數倍數
		a3=b/2-a/2;//區間一中2的倍數的個數
		if(a%2==0)//注意！！
		{
			a3++;
		}
		b3=d/2-c/2;//區間二中2的倍數的個數
		if(c%2==0)//注意！！
		{
			b3++;
		}
		sum=sum+b2*(a3-a1)+a2*(b3-b1);//前面的2018倍數個數+區間一1009的奇數倍個數*區間二2的倍數-2018倍數+區間二1009的奇數倍個數*區間一2的倍數-2018倍數
		printf("%lld\n",sum);
	}
	return 0;
}