题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6286
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.
## Constraint
* 1≤a≤b≤10^9,1≤c≤d≤10^9
* The number of tests cases does not exceed 10^4.
正解：容斥原理：
因为2018的因子只有1、2、1009、2018，所以只需计算以下五种情况：
1. [a,b]中2018的倍数，[c,d]为任意数
2. [c,d]中2018的倍数，[a,b]为任意数
3. [a,b]中2018的倍数且[c,d]中2018的倍数（为了1，2情况去重）
4. [a,b]中1009的奇数倍（偶数倍同1有重叠），[d,c]中2的倍数且不是2018的倍数
5. [c,d]中1009的奇数倍（偶数倍同2有重叠），[a,b]中2的倍数且不是2018的倍数
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<vector> 5 #include<queue> 6 #include<algorithm> 7 using namespace std; 8 typedef long long ll; 9 ll a, b, c, d; 10 ll solve(ll x) 11 { 12 x /= 1009; 13 if (x % 2) return x / 2 + 1; 14 return x / 2; 15 } 16 int main() 17 { 18 while (~scanf("%lld%lld%lld%lld", &a, &b, &c, &d)) { 19 ll sum = 0; 20 sum += (b / 2018 - (a - 1) / 2018)*(d - c + 1); 21 sum += (d / 2018 - (c - 1) / 2018)*(b - a + 1); 22 sum -= (b / 2018 - (a - 1) / 2018)*(d / 2018 - (c - 1) / 2018); 23 sum += (solve(b) - solve(a - 1))*(d / 2 - (c - 1) / 2 - (d / 2018 - (c - 1) / 2018)); 24 sum += (solve(d) - solve(c - 1))*(b / 2 - (a - 1) / 2 - (b / 2018 - (a - 1) / 2018)); 25 printf("%lld\n", sum); 26 } 27 return 0; 28 }
非常规解法：
先暴力找出[0，2017]范围内所有满足i*j%2018=0的数对(i，j)，这样的数对数目不超过6050个，
又有(i*j)%p=[(i%p)*(j%p)]%p，在这里i%p、j%p的值必然小于2018，若它们乘起来可被2018整除，就说明这样的i、j满足要求。
而我们又找出了所有2018内的数对(i，j)，对于每一个(i，j)，我们求出[a,b]间被2018整除余i的数字个数，乘以[c,d]间被2018整除余j的数字个数，将其累加起来即为答案。
补充：vjudge上没法直接交表，所以这里不得不把打表的过程写进去，导致编译器要选g++才可以过，c++会超时。
1 #pragma GCC optimize (3) 2 #include<vector> 3 #include<stack> 4 #include<bitset> 5 #include<cstdlib> 6 #include<cmath> 7 #include<set> 8 #include<list> 9 #include<deque> 10 #include<map> 11 #include<queue> 12 #include<iostream> 13 #include<cstdio> 14 #include<cstring> 15 #include<algorithm> 16 #define ull unsigned long long 17 using namespace std; 18 int ar[6051][2]; 19 inline int solve(int r, int chu, int mod) 20 { 21 if (r <= 0) return 0; 22 if (mod == 0) return r / chu; 23 if (r < mod) return 0; 24 if ((r - mod + 1) % chu == 0) return (r - mod + 1) / chu; 25 else return (r - mod + 1) / chu + 1; 26 } 27 int main() 28 { 29 ull ans; 30 int a, b, c, d, k = 0; 31 for (int i = 0; i <= 2017; i++) { 32 for (int j = 0; j <= 2017; j++) { 33 if ((i*j) % 2018 == 0) { 34 ar[k][0] = i; 35 ar[k][1] = j; 36 k++; 37 } 38 } 39 } 40 while (scanf("%d%d%d%d", &a, &b, &c, &d) == 4) { 41 ans = 0; 42 for (int i = 0; i <= 6050; i++) { 43 ans += (ull)(solve(b, 2018, ar[i][0]) - solve(a - 1, 2018, ar[i][0]))*(solve(d, 2018, ar[i][1]) - solve(c - 1, 2018, ar[i][1])); 44 } 45 cout << ans << endl; 46 } 47 return 0; 48 }