HDU - 3709 - Balanced Number (digital DP)

link:

https://vjudge.net/problem/HDU-3709

Meaning of the questions:

A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. More specifically, imagine each digit as a box with weight indicated by the digit. When a pivot is placed at some digit of the number, the distance from a digit to the pivot is the offset between it and the pivot. Then the torques of left part and right part can be calculated. It is balanced if they are the same. A balanced number must be balanced with the pivot at some of its digits. For example, 4139 is a balanced number with pivot fixed at 3. The torqueses are 42 + 11 = 9 and 9*1 = 9, for left part and right part, respectively. It's your job
to calculate the number of balanced numbers in a given range [x, y].

Ideas:

At first I did not count specific size. . That would be considered direct violence space.
And record the current moment, so high is a positive number, so from high to low when a negative value if the torque side, you can directly return 0, because the lower weights are negative.
At the same time position to enumerate each case as a fulcrum, subtracting each count 0 again.

Code:

// #include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<vector>
#include<string.h>
#include<set>
#include<queue>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long LL;
const int MOD = 1e9+7;
const int MAXN = 1e6+10;

LL F[20][20][1600];
LL dig[20];
LL a, b;

LL Dfs(int pos, int piv, int sum, bool lim)
{
    if (pos == -1)
        return sum == 0;
    if (sum < 0)
        return 0;
    if (!lim && F[pos][piv][sum] != -1)
        return F[pos][piv][sum];
    int up = lim ? dig[pos] : 9;
    LL ans = 0;
    for (int i = 0;i <= up;i++)
        ans += Dfs(pos-1, piv, (pos-piv)*i+sum, lim && i == up);
    if (!lim)
        F[pos][piv][sum] = ans;
    return ans;
}

LL Solve(LL x)
{
    int p = 0;
    while(x)
    {
        dig[p++] = x%10;
        x /= 10;
    }
    LL ans = 0;
    for (int i = 0;i < p;i++)
        ans += Dfs(p-1, i, 0, true);
    return ans-(p-1);
}

int main()
{
    // freopen("test.in", "r", stdin);
    memset(F, -1, sizeof(F));
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%lld%lld", &a, &b);
        printf("%lld\n", Solve(b)-Solve(a-1));
    }

    return 0;
}