2018(容斥定理 HDU6286)

Posted 2021-02-05 chr1stopher

2018

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 507    Accepted Submission(s): 263

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

思路：

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

AC代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<string>
#include<queue>
#include<utility>
#define ll long long
using namespace std;
const int maxn = 1e5+7;

int main(int argc, char const *argv[])
{
    ll a,b,c,d;
    while(~scanf("%lld %lld %lld %lld",&a,&b,&c,&d))
    {
        ll s1 = b-a+1;
        ll s2 = d-c+1;
        ll s1_odd = b/2-(a-1)/2;
        ll s1_1009 = b/1009-(a-1)/1009;

        ll x1=(b/2018-(a-1)/2018)*s2;
        ll x2=(s1_odd-(b/2018-(a-1)/2018))*(d/1009-(c-1)/1009);
        ll x3=(s1_1009-(b/2018-(a-1)/2018))*(d/2-(c-1)/2);
        ll x4=((s1-s1_odd)-(s1_1009-(b/2018-(a-1)/2018)))*(d/2018-(c-1)/2018);
        printf("%lld
",x1+x2+x3+x4);
    }
    return 0;
}