題解

題是真的好，我是真的不會做
智商本還是要多開啊QwQ

我們發現一個非下降的數字一定可以用不超過九個1111111111...1111表示

那麽我們可以得到這樣的一個式子，假如我們用了k個數，那麽最多的話可以是這樣的
\(N = \sum_{i = 1}^{9k} (10^{r_i} - 1) / 9\)
\(9N + 9k = \sum_{i = 1}^{9k} 10^{r_{i}}\)

我們只要每次計算出9N + 9 ，9N + 18...，然後看看十進制下每一位的數字和有沒有超過9k，直接加的話最壞情況是一次操作\(O(L)\)的，但是大家應該都有種直覺總的操作就是\(O(L)\)的……就是勢能分析啦，不太會證，就是一次長的進位過後之後不會再進位了。。。

復雜度\(O(lg N)\)

代碼

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
#include <cmath>
#include <bitset>
#include <queue>
#define enter putchar(‘\n‘)
#define space putchar(‘ ‘)
//#define ivorysi
 

#define pb push_back
#define mo 974711
#define pii pair<int,int>
#define mp make_pair
#define fi first
#define se second
#define MAXN 200005
#define eps 1e-12
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1
 
;
    while(c < ‘0‘ || c > ‘9‘) {
    if(c == ‘-‘) f = -1;
    c = getchar();
    }
    while(c >= ‘0‘ && c <= ‘9‘) {
    res = res * 10 - ‘0‘ + c;
    c = getchar();
    }
    res = res * f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar(‘-‘);}
    if(x >= 10) out(x / 10);
    putchar(‘0‘ + x % 10);
}
struct Bignum {
    vector<int> v;
    int sum;
    Bignum operator = (string s) {
    v.clear();
    sum = 0;
    for(int i = s.length() - 1 ; i >= 0 ; --i) {
        v.pb(s[i] - ‘0‘);
        sum += s[i] - ‘0‘;
    }
    return *this;
    }
    friend Bignum operator * (const Bignum &a,const int b) {
    int s = a.v.size();
    Bignum c;c.v.clear();
    for(int i = 0 ; i <= s ; ++i) c.v.pb(0);
    int g = 0;
    for(int i = 0 ; i < s ; ++i) {
        int x = a.v[i] * b + g;
        c.v[i] = x % 10;
        g = x / 10;
    }
    if(g) c.v[s] = g;
    for(int i = s ; i > 0 ; --i) {
        if(c.v[i] == 0) c.v.pop_back();
        else break;
    }
    c.sum = 0;s = c.v.size();
    for(int i = 0 ; i < s ; ++i) c.sum += c.v[i];
    return c;
    }
}A;
string s;
void Solve() {
    cin>>s;
    A = s;
    A = A * 9;
    int ans = 0;
    while(1) {
    int s = A.v.size();
    int g = 9;
    for(int i = 0 ; i < s ; ++i) {
        if(!g) break;
        A.sum -= A.v[i];
        int x = A.v[i] + g;
        A.v[i] = x % 10;
        A.sum += A.v[i];
        g = x / 10;
    }
    if(g) A.v.pb(g),A.sum += g;
    ++ans;
    if(ans * 9 >= A.sum) break;
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

【AtCoder】AGC011 E - Increasing Numbers