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题意：给定两个区间 [a， b]， [c， d]，问有多少个有序对 (x, y)（x >= a && x <= b && y >= c && y <= d）使得，（x*y）= 0（mod 2018）。
输入：a， b， c， d。
数据范围：多组输入，所有输入均为[1, 1e9]。
The number of tests cases does not exceed 104104. 

/*
a b为左区间，c d为右区间
先求左区间中2018的个数，乘右区间的个数，同理求右区间
再求左区间中1009的倍数，但不是2018的倍数，再求右区间中2的倍数，但不是2018的倍数，同理右区间
再减去左区间和右区间中都是2018的倍数

*/
#include<bits/stdc++.h>

using namespace std;
typedef unsigned long long ll;

int main()
{
    ll a,b,c,d;
    while(scanf("%llu%llu%llu%llu",&a,&b,&c,&d)==4)
    {
        ll tip1=a/2018,tip2=b/2018;
        ll x1=tip2-tip1;
        if(a%2018==0 )
            x1++;

        tip1=c/2018,tip2=d/2018;
        ll x2=tip2-tip1;
        if(c%2018==0 )
            x2++;


        tip1=a/1009,tip2=b/1009;
        ll y1=tip2-tip1;
        if(a%1009==0 )
            y1++;
        y1-=x1;

        tip1=c/1009,tip2=d/1009;
        ll y2=tip2-tip1;
        if(c%1009==0 )
            y2++;
        y2-=x2;

        tip1=a/2,tip2=b/2;
        ll z1=tip2-tip1;
        if(a%2==0)
            z1++;
        z1-=x1;

        tip1=c/2,tip2=d/2;
        ll z2=tip2-tip1;
        if(c%2==0 )
            z2++;
        z2-=x2;
       // cout << x1 << " " << x2 << " " << y1 << " " << y2 << " " << z1 << " " << z2 << endl;

        ll ans=(b-a+1)*x2 + (d-c+1)*x1 + (y1*z2) + (y2*z1)-x1*x2;
        cout << ans << endl;
    }
    return 0;
}



或者
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
int main()
{
    LL a,b,c,d;
    while(scanf("%lld%lld%lld%lld",&a,&b,&c,&d)==4)
    {
        LL ans=0;
        LL x=0,y=0;
        x=(b/1009-(a-1)/1009);
        y=(b/2018-(a-1)/2018);
        x-=y;
        ans+=((b+1)/2-a/2-x)*(d/2018-(c-1)/2018);
        ans+=(b/2-(a-1)/2-y)*(d/1009-(c-1)/1009);
        ans+=y*(d-c+1);
        ans+=x*(d/2-(c-1)/2);
        printf("%lld
",ans);
    }
}