题目大意：给据题目定义，找到l，r，使得solve(l,r) % a = 0.
解题思路：f(x + 1e18) = f(x) + 1,所以有solve(x + 1, x+1e18) = solve(x, x+1e18-1) + 1，假定x为0，我们求出solve(0, 1e18) % a = k，那么a - k，即为区间需要移动的步长。solve(1e18) % a = 4518
1e17 % a
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
ll A, M = 1e17;
int main () {
scanf("%lld", &A);
A -= 5 * (9 * (M * 18 % A) % A) % A;
printf("%lld %lld\n", A, A + M * 10 - 1);
return 0;
}