题目链接：Codeforces 468C Hack it!

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 = 45181e17 % 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;
}

Codeforces 468C Hack it!(数学)