F

题目链接：https://vjudge.net/contest/365059#problem/F

题目大意：给定区间[a,b]，求区间内平衡数的个数。所谓平衡数即有一位做平衡点，左右两边数字的力矩想等。

想法：

枚举起点的位置去进行数位DP

#pragma GCC optimize(3,"Ofast","inline")//O3优化
#pragma GCC optimize(2)//O2优化
#include <algorithm>
#include <string>
#include <string.h>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <queue>
#include <math.h>
#include <cstdio>
#include <iomanip>
#include <time.h>
#include <bitset>
#include <cmath>
#include <sstream>
#include <iostream>
#include <cstring>

#define LL long long
#define ls nod<<1
#define rs (nod<<1)+1
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define INF 0x3f3f3f3f
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)

const double eps = 1e-10;
const int maxn = 2e5 + 10;
const int mod = 1e9 + 7;

int sgn(double a){return a < -eps ? -1 : a < eps ? 0 : 1;}
using namespace std;

LL L,R;
int len;
int b[20];
LL mem[20][25][2000];

LL dfs(int cur,int pos,int tmp,bool f) {
    if (cur < 0)
        return tmp == 0;
    if (tmp < 0)
        return 0;
    if (!f && mem[cur][pos][tmp] != -1)
        return mem[cur][pos][tmp];
    int v = 9;
    if (f)
        v = b[cur];
    LL ans = 0;
    for (int i = 0;i <= v;i++) {
        ans += dfs(cur-1,pos, tmp + (cur-pos)*i,f && (i == v));
    }
    if (!f)
        mem[cur][pos][tmp] = ans;
    return ans;
}

LL solve(LL x) {
    if (x < 0)
        return 0;
    len = 0;
    while (x) {
        b[len++] = x % 10;
        x /= 10;
    }
    LL ans = 0;
    for (int i = 0;i < len;i++) {
        ans += dfs(len-1,i,0,1);
    }
    return ans-len+1;
}

int main() {
    ios::sync_with_stdio(0);
    int T;
    cin >> T;
    memset(mem,-1, sizeof(mem));
    while (T--) {
        cin >> L >> R;
        cout << solve(R) - solve(L-1) << endl;
    }
    return 0;
}