cf Educational Codeforces Round 50 C. Classy Numbers

原題：
C. Classy Numbers
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Let’s call some positive integer classy if its decimal representation contains no more than 3 non-zero digits. For example, numbers 4, 200000, 10203 are classy and numbers 4231, 102306, 7277420000 are not.You are given a segment [L;R]
. Count the number of classy integers x such that L≤x≤R.

Each testcase contains several segments, for each of them you are required to solve the problem separately.

Input
The first line contains a single integer T (1≤T≤10^4) — the number of segments in a test case.

Each of the next T lines contains two integers Li and Ri (1≤Li≤Ri≤10^18).

Output
Print T lines — the i-th line should contain the number of classy integers on a segment [Li;Ri].

Example
input
4
1 1000
1024 1024
65536 65536
999999 1000001
output
1000
1
0
2

中文：

給你一個區間，讓你判斷這個區間當中整數中非零位不超過3個的數字的個數。

代碼：

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

vector<ll> res;
int t;

void dfs(int pos,int cnt,ll cur)
{
    if(18==pos)
    {
        res.push_back(cur);
        return;
    }
    dfs(pos+1,cnt,cur*10);
    if(cnt<3)
    {
        for(ll i=1;i<=9;i++)
            dfs(pos+1,cnt+1,cur*10+i);
    }
}

int main() 
{
	ios::sync_with_stdio(false);
    dfs(0, 0, 0);
    res.push_back(1000000000000000000);

    cin>>t;
    while(t--)
    {
        ll L, R;
        cin>>L>>R;
        cout<<upper_bound(res.begin(), res.end(), R) - lower_bound(res.begin(), res.end(), L)<<endl;
    }
    return 0;
}

解答：

這是純的排列組合計數問題，大概思路是，可以先通過算出整10次冪之內有多少個滿足條件的數，然後從左向右，每次在後面的位數當中選3位置、選2個位置、選1個位置以及選0個位置添加非0位，自己推了個計數公式，算的漏洞百出。-_-!!!

官方題解中給出了一種特別輕鬆的解決方法，就是暴力把所有數枚舉一遍，然後使用二分查找兩個區間端點即可，太簡單了，可惜自己當時就顧着算公式，沒想到這辦法。