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

注释里写了 就是看左右区间多少个满足条件的数乘起来

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
typedef long long ll;
using namespace std;
//2018的因数只有2018 1009 2 1
//小技巧 用6来代替2018便于理解
//左区间个数 乘 右区间个数
//6的倍数 1的倍数（整个区间） 6的倍数乘以任何数都是6的倍数
//3的倍数（除去6的倍数） 2的倍数
//2的倍数（除去6的倍数） 3的倍数
//剩下的数 6的倍数
void test(){
    for(int i=1;i*i<=2018;i++){
        if(2018%i==0){
            printf("%d\n",i);
            if(2018/i!=i){
                printf("%d\n",2018/i);
            }
        }
    }
}
int main(void){
    //test();
    ll a,b,c,d;
    while(~scanf("%lld%lld%lld%lld",&a,&b,&c,&d)){
        ll ans=0;
        //左6的倍数
        ans+=((b/2018)-((a-1)/2018))*((d-c)+1);
        //左3的倍数
        ans+=((((b/1009)-((a-1)/1009))-((b/2018)-((a-1)/2018)))*((d/2)-((c-1)/2)));
        //左2的倍数
        ans+=((((b/2)-((a-1)/2))-((b/2018)-((a-1)/2018)))*((d/1009)-((c-1)/1009)));
        //剩余的数
        ans+=(b-a+1-(((b/1009)-((a-1)/1009))-((b/2018)-((a-1)/2018)))-(((b/2)-((a-1)/2))-((b/2018)-((a-1)/2018)))-((b/2018)-((a-1)/2018)))*(((d/2018)-((c-1)/2018)));
        printf("%lld\n",ans); 
    }
    return 0;
}