2018ACM-CCPC湖南湘潭邀请赛 2018 容斥原理

最新推荐文章于 2021-07-09 13:35:45 发布

Cherry_0525

最新推荐文章于 2021-07-09 13:35:45 发布

阅读量398

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/qq_41037114/article/details/80378193

本文介绍了一种算法问题，给定四个整数a、b、c、d，求解在区间[a, b]和[c, d]内能够构成乘积为2018倍数的所有整数对(x, y)的数量。通过容斥原理分解问题，并提供了一个具体的C++实现示例。

2018

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

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 20181 2018 1 20181 1000000000 1 1000000000

Sample Output

360511485883320325200

//给你a、b、c、d,其中x属于a-b，y属于c-d，问你有几种x，y的可能
//容斥原理
//1、x是偶数也是1009的倍数，y可以是所有数；2、x是偶数但不是1009的倍数，y是1009的倍数
//3、x是奇数也是1009的倍数，y是偶数；      3、x是奇数但不是1009的倍数，y是2018的倍数

#include <bits/stdc++.h>
#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;
        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\n",x1+x2+x3+x4);
    }
    return 0;
}