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.
The input consists of several test cases and is terminated by end-of-file. Each test case contains four integers a,b,c,d.
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.
1 2 1 2018
1 2018 1 2018
1 1000000000 1 1000000000
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 xy 是 2018 的倍数 思路:2018=21009(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:
后 AC 代码
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 | #include<cstdio> #include<iostream> typedef unsigned long long ll; using namespace std; int main(){ ll a,b,c,d; while(cin>>a>>b>>c>>d){ ll num1_all_1009=b/1009-(a-1)/1009; ll num1_even=b/2-(a-1)/2; ll num1_1009_in_even=b/2018-(a-1)/2018; ll num1_rest_in_even=num1_even-num1_1009_in_even; ll num1_odd=(b-a+1)-num1_even; ll num1_1009_in_odd=num1_all_1009-num1_1009_in_even; ll num1_rest_in_odd=num1_odd-num1_1009_in_odd; ll ans=0; ans+=num1_1009_in_even*(d-c+1); ll num2_all_1009=d/1009-(c-1)/1009; ans+=num1_rest_in_even*num2_all_1009; ll num2_even=d/2-(c-1)/2; ans+=num1_1009_in_odd*num2_even; ll num2_all_2018=d/2018-(c-1)/2018; ans+=num1_rest_in_odd*num2_all_2018; cout<<ans<<endl; } return 0; } |
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