HDU 6659 Acesrc and Good Numbers (数学 思维)

2019 杭电多校 8 1003

题目链接：HDU 6659

比赛链接：2019 Multi-University Training Contest 8

Problem Description

Acesrc is a famous mathematician at Nanjing University second to none. Playing with interesting numbers is his favorite. Today, he finds a manuscript when cleaning his room, which reads

... Let \(f(d,n)\) denote the number of occurrences of digit \(d\) in decimal representations of integers \(1,2,3,⋯,n\). The function has some fantastic properties ...

... Obviously, there exist some nonnegative integers \(k\), such that \(f(d,k)=k\), and I decide to call them \(d\)-good numbers ...

... I have found all d-good numbers not exceeding \(10^{1000}\), but the paper is too small to write all these numbers ...

Acesrc quickly recollects all \(d\)-good numbers he found, and he tells Redsun a question about \(d\)-good numbers: what is the maximum \(d\)-good number no greater than \(x\)? However, Redsun is not good at mathematics, so he wants you to help him solve this problem.

Input

The first line of input consists of a single integer \(q (1\le q\le 1500)\), denoting the number of test cases. Each test case is a single line of two integers \(d (1\le d\le 9)\) and \(x (0\le x\le 10^{18})\).

Output

For each test case, print the answer as a single integer in one line. Note that \(0\) is trivially a \(d\)-good number for arbitrary \(d\).

Sample Input

3
1 1
1 199999
3 0

Sample Output

1
199990
0

Solution

题意

定义 \(f(d, n)\) 为十进制下 \(1\) 到 \(n\) 所有数的数位中数字 \(d\) 出现的次数。给定 \(x\)，找出最大的 \(n(n \le x)\) 满足 \(f(d, n) = n\)。

题解

看到了一个神仙做法。

显然如果 \(f(d, x) = x\) 时就直接输出。

否则，需要缩小 \(x\)。令 \(f(d, x) = y\)，则需要将 \(x\) 缩小 \(\lceil \frac{|x - y|}{18} \rceil\)。即 \(x = x - abs(f(d, x) - x) / 18\)。原因是 \(f(d, x)\) 与 \(f(d, x - 1)\) 最多相差 \(18\) 个 \(d\) \(\ (e.g. \ f(9, 10^{18}-1)\ to\ f(9, 10^{18}-2))\)。

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

// 计算 1 到 n 中数字 x 出现的次数
ll f(ll d, ll n) {
    ll cnt = 0, k;
    for (ll i = 1; k = n / i; i *= 10) {
        cnt += (k / 10) * i;
        int cur = k % 10;
        if (cur > d) {
            cnt += i;
        }
        else if (cur == d) {
            cnt += n - k * i + 1;
        }
    }
    return cnt;
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int T;
    cin >> T;
    while (T--) {
        ll d, x;
        cin >> d >> x;
        while (true) {
            ll num = f(d, x);
            if (num == x) {
                cout << x << endl;
                break;
            } else {
                x -= max(1LL, abs(num - x) / 18);
            }
        }
    }
    return 0;
}

Reference

2019 Multi-University Training Contest 8——Acesrc and Good Numbers（数学 想法）