CF1036C Classy Numbers dfs+二分

Classy Numbers

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's call some positive integer classy if its decimal representation contains no more than 33 non-zero digits. For example, numbers 44, 200000200000, 1020310203 are classy and numbers 42314231, 102306102306, 72774200007277420000 are not.

You are given a segment [L;R][L;R]. Count the number of classy integers xx such that L≤x≤RL≤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 TT (1≤T≤1041≤T≤104) — the number of segments in a testcase.

Each of the next TT lines contains two integers LiLi and RiRi (1≤Li≤Ri≤10181≤Li≤Ri≤1018).

Output

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

Example

input

Copy

4
1 1000
1024 1024
65536 65536
999999 1000001

output

Copy

1000
1
0
2

题意：classy数：一个数的不为0的位数不超过三位，问你le到ri范围内有多少个classy数？

分析：考虑le和ri的范围为10^18，这之间满足条件的classy数不超过一百万（自己暴搜计算一下），将这些数存进数组里，排下序找出le的位置和ri的位置，相减就可以得出中间classy数的多少

AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e5+10;
const ll mod = 2e9+7;
const double pi = acos(-1.0);
const double eps = 1e-8;
vector<ll> v;
void dfs( ll dep, ll res, ll n ) { //dep表示现在这个数是第几位，res表示现在这个数的大小，n表示res中不为0的位数
    v.push_back(res);
    if( dep == 18 ) {
        return ;
    }
    dfs(dep+1,res*10,n);
    if( n < 3 ) {
        for( ll i = 1; i <= 9; i ++ ) {
            dfs(dep+1,res*10+i,n+1);
        }
    }
}
int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    ll T, le, ri;
    for( ll i = 1; i <= 9; i ++ ) {
        dfs(1,i,1);
    }
    v.push_back(1e18);
    sort(v.begin(),v.end());  //排序数组
    //debug(v.size());
    cin >> T;
    while( T -- ) {
        cin >> le >> ri;
        cout << upper_bound(v.begin(),v.end(),ri)-lower_bound(v.begin(),v.end(),le) << endl;  //二分查找
    }
    return 0;
}