CCPC2018-湖南全国邀请赛 K 2018

K.2018

题目描述

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.

输入

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

输出

For each test case, print an integer which denotes the result.
• 1 ≤ a ≤ b ≤ 10⁹ , 1 ≤ c ≤ d ≤ 10⁹
• The number of tests cases does not exceed 10⁴ .

样例输入

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

样例输出

3
6051
1485883320325200

 

题意：给定区间[a,b]、[c,d]，问有多少对有序数组(x,y)(x∈[a,b],y∈[c,d])使得x*y是2018的倍数
思路：2018=2*1009(分解质因数)，则对x分类讨论:1)仅为2的倍数；2）仅为1009的倍数；3）即为2又为1009的倍数；4）既不为2又不为1009的倍数
等价于如下分类讨论：
1.若x是偶数：1）若x是1009的倍数，则y可为[c,d]中任意数； 2）若x不是1009的倍数，则y必定为[c,d]中1009的倍数
2.若x是奇数：1）若x是1009的倍数，则y必定为[c,d]中2的倍数； 2）若x不是1009的倍数，则y必定为[c,d]中2018的倍数

#include <iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define ll long long
using namespace std;
int main()
{
    ll a,b,c,d;
    while(cin>>a>>b>>c>>d)
    {
        ll s1=b-a+1;
        ll s2=d-c+1;
        ll s1_o=b/2-(a-1)/2;
        ll s1_1009=b/1009-(a-1)/1009;

        /*int x=b/2018-(a-1)/2018;
        ans+=x*s2;

        x=s1_o-x;
        ans+=x*(d/1009-(c-1)/1009);

        x=s1_1009-(b/2018-(a-1)/2018);
        ans+=x*(d/2-(c-1)/2);

        x=s1-s1_o-x;
        ans+=x*(d/2018-(c-1)/2018);*/
        ll x1,x2,x3,x4;

        x1=(b/2018-(a-1)/2018)*s2;
        x2=(s1_o-(b/2018-(a-1)/2018))*(d/1009-(c-1)/1009);
        x3=(s1_1009-(b/2018-(a-1)/2018))*(d/2-(c-1)/2);
        x4=((s1-s1_o)-(s1_1009-(b/2018-(a-1)/2018)))*(d/2018-(c-1)/2018);
        printf("%lld
",x1+x2+x3+x4);
    }
    return 0;
}