Hack It!


Little X has met the following problem recently.
Let’s define $f(x)$ as the sum of digits in decimal representation of number $x$ (for example, $f(1234)=1+2+3+4$). You are to calculate $\sum_{i=l}^r f(i) \bmod a$.
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:

ans = solve(l, r) % a;
if (ans <= 0) ans += a;

This code will fail only on the test with $\sum_{i=l}^r f(i) \equiv 0 \pmod a$. You are given number $a$, help Little X to find a proper test for hack.


令$f(i)$表示$i$各位数字之和。给定$a$,要求找出一对$(l, r)$满足$\sum_{i=l}^r f(i) \equiv 0 \pmod a$。
数据范围:$1 \le a \le 10^{18}$。


思路很巧妙。有一个显而易见的结论:$\forall i \in [1, 10^{18}], \; f(i)+1=f(i+10^{18})$。它有一个推论:$\forall x \in [1, 10^{18}], \; 1+\sum_{i=x}^{x+10^{18}-1} f(i)=\sum_{i=x+1}^{x+10^{18}} f(i)$。因此只要算出$t=\sum_{i=1}^{10^{18}} f(i) \bmod a$,将区间右移$a-t$单位。


#include <cstdio>

long long mul(long long a, long long b, long long mod) {
  long long ret = 0;
  while (b) b & 1 && ((ret += a) %= mod), (a <<= 1) %= mod, b >>= 1;
  return ret;

int main() {
  long long n; scanf("%lld", &n);
  long long cur = mul(1800000000000000000ll, 45, n) + 1, l = 1, r = 1000000000000000000ll;
  l += n - cur, r += n - cur, printf("%lld %lld\n", l, r);
  return 0;

RegMs If

418 I'm a teapot

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