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 testcase.

Each of the next T lines contains two integers L i and R i ( 1 ≤ L i ≤ R i ≤ 10 18 ).

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

Example inputCopy 4 1 1000 1024 1024 65536 65536 999999 1000001 outputCopy 1000 1 0 2 solve:數位dp即可，記憶化一下,dfs傳入三個變數:pos(數位)，num(當前還差幾個數就能達到3個非0數),limit(是否到達上界)。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+10;
const int INF = 0x3f3f3f3f;
typedef long long ll;
typedef int INT ;
#define int long long
int digit[20];
int dp[20][30];
int dfs(int pos,int num,int limit){
    if(pos<0&&num>=0) return 1;
    if(pos<0||num<0) return 0;
    if(!limit&&dp[pos][num]) return dp[pos][num];

    int ans=0;
    int up=limit?digit[pos]:9;

    for(int i=0;i<=up;i++){
        ans+=dfs(pos-1,num-(i?1:0),limit&&i==up);
    } 
    if(!limit) dp[pos][num]=ans;
    return ans;
}

int solve(int x){
    int cnt=0;
    while(x){
        digit[cnt++]=x%10;
        x/=10;
    }
    return dfs(cnt-1,3,1);
}

INT main(){
    int t,a,b;
    cin>>t;
    while(t--){
        memset(dp,0,sizeof(dp));
        cin>>a>>b;
        cout<<solve(b)-solve(a-1)<<endl;
    }
    return 0;
}