PROBLEM LINK:
DIFFICULTY:
MEDIUM
PREREQUISITES:
SEARCHING ALGORITHMS
EXPLANATION:
If x’s length is 1, then the answer is 0 or 1. Let us consider otherwise.
Let f(n) be the value of X interpreted base n, then f is strictly monotonically increasing.
Therefore, one can find the maximum n such that f(n)≤M with binary search, and thus solve the problem.
When X is 10, the answer can be at most about M.
Be careful of overflows in the course of calculation.
Let L be the length of X, then the time complexity is O(LlogM). With proper implementation, one can solve it in an O(LogM) time (plus receiving the input).
SOLUTION:
#include <bits/stdc++.h>
using namespace std;
// clang-format off
// template {{{
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define f first
#define s second
#define sz(x) int((x).size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define trav(a, x) for (auto &a : x)
#define L1(u, ...) [&](auto &&u) { return __VA_ARGS__; }
#define L2(u, v, ...) [&](auto &&u, auto &&v) { return __VA_ARGS__; }
#define sort_by(x, y) sort(all(x), [&](const auto &l, const auto &r) { return y; })
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vvb = vector<vb>;
using vd = vector<double>;
using vs = vector<string>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pdd = pair<double, double>;
using vpii = vector<pii>;
using vpll = vector<pll>;
using vpdd = vector<pdd>;
template <typename T> void ckmin(T &a, const T &b) { a = min(a, b); }
template <typename T> void ckmax(T &a, const T &b) { a = max(a, b); }
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
namespace __input {
template <class T1, class T2> void re(pair<T1, T2> &p);
template <class T> void re(vector<T> &a);
template <class T, size_t SZ> void re(array<T, SZ> &a);
template <class T> void re(T &x) { cin >> x; }
void re(double &x) { string t; re(t); x = stod(t); }
template <class Arg, class... Args> void re(Arg &first, Args &...rest) { re(first); re(rest...); }
template <class T1, class T2> void re(pair<T1, T2> &p) { re(p.f, p.s); }
template <class T> void re(vector<T> &a) { for (int i = 0; i < sz(a); i++) re(a[i]); }
template <class T, size_t SZ> void re(array<T, SZ> &a) { for (int i = 0; i < SZ; i++) re(a[i]); }
}
using namespace __input;
namespace __output {
template <typename T> struct is_outputtable { template <typename C> static constexpr decltype(declval<ostream &>() << declval<const C &>(), bool()) test(int) { return true; } template <typename C> static constexpr bool test(...) { return false; } static constexpr bool value = test<T>(int()); };
template <class T, typename V = decltype(declval<const T &>().begin()), typename S = typename enable_if<!is_outputtable<T>::value, bool>::type> void pr(const T &x);
template <class T, typename V = decltype(declval<ostream &>() << declval<const T &>())> void pr(const T &x) { cout << x; }
template <class T1, class T2> void pr(const pair<T1, T2> &x);
template <class Arg, class... Args> void pr(const Arg &first, const Args &...rest) { pr(first); pr(rest...); }
template <class T, bool pretty = true> void prContain(const T &x) { if (pretty) pr("{"); bool fst = 1; for (const auto &a : x) pr(!fst ? pretty ? ", " : " " : "", a), fst = 0; if (pretty) pr("}"); }
template <class T> void pc(const T &x) { prContain<T, false>(x); pr("\n"); }
template <class T1, class T2> void pr(const pair<T1, T2> &x) { pr("{", x.f, ", ", x.s, "}"); }
template <class T, typename V, typename S> void pr(const T &x) { prContain(x); }
void ps() { pr("\n"); }
template <class Arg> void ps(const Arg &first) { pr(first); ps(); }
template <class Arg, class... Args> void ps(const Arg &first, const Args &...rest) { pr(first, " "); ps(rest...); }
}
using namespace __output;
#define __pn(x) pr(#x, " = ")
#ifdef ANAND_LOCAL
#define pd(...) pr("\033[1;31m"), __pn((__VA_ARGS__)), ps(__VA_ARGS__), pr("\033[0m"), cout << flush
#else
#define pd(...)
#endif
namespace __algorithm {
template <typename T> void dedup(vector<T> &v) { sort(all(v)); v.erase(unique(all(v)), v.end()); }
template <typename T> typename vector<T>::const_iterator find(const vector<T> &v, const T &x) { auto it = lower_bound(all(v), x); return it != v.end() && *it == x ? it : v.end(); }
template <typename T> size_t index(const vector<T> &v, const T &x) { auto it = find(v, x); assert(it != v.end() && *it == x); return it - v.begin(); }
template <typename I> struct _reversed_struct { I &v_; explicit _reversed_struct(I &v) : v_{v} {} typename I::reverse_iterator begin() const { return v_.rbegin(); } typename I::reverse_iterator end() const { return v_.rend(); } };
template <typename I> _reversed_struct<I> reversed(I &v) { return _reversed_struct<I>(v); }
}
using namespace __algorithm;
namespace __io {
void setIO() { ios_base::sync_with_stdio(0); cin.tie(0); cout << setprecision(15); }
}
using namespace __io;
// }}}
// clang-format on
// Returns first i in [l, r] s.t. predicate(i) is true. Never evaluates r.
template <typename I, typename P> I binarysearch(const P &predicate, I l, I r) {
l--;
while (r - l > 1) {
auto mid = l + (r - l) / 2;
if (predicate(mid))
r = mid;
else
l = mid;
}
return r;
}
int main() {
setIO();
string input;
re(input);
vi x;
for (char c : input)
x.push_back(c - '0');
ll m;
re(m);
if (sz(x) == 1) {
if (x.front() > m)
ps(0);
else
ps(1);
return 0;
}
auto test = [&](ll base) -> ll {
ll res = 0;
for (int i : x) {
if (res > (m - i) / base)
return true;
res = res * base + i;
if (res > m)
return true;
}
return false;
};
int lo = *max_element(all(x)) + 1;
ll ans = -lo + binarysearch(test, 1LL, m + 1);
ans = max(ans, 0LL);
ps(ans);
return 0;
}