Balanced Number

Problem Description
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 4*2 + 1*1 = 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].
 Input
The input contains multiple test cases. The first line is the total number of cases T (0 < T ≤ 30). For each case, there are two integers separated by a space in a line, x and y. (0 ≤ x ≤ y ≤ 10 18).
 Output
For each case, print the number of balanced numbers in the range [x, y] in a line.
 Sample Input
2 0 9 7604 24324
 Sample Output
10 897
 

判断一个数是不是“平衡的”
枚举支点，算力矩……力矩为负的时候可以直接剪枝…… 
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
long long a, b;
long long f[30][30][2005];
int bit[30];
long long dp(int pos, int x, int st, int flag)   //st是力矩
{
    if (pos == 0) return st == 0;
    if (st < 0) return 0;
    if (flag && f[pos][x][st] != -1) return f[pos][x][st];
    int d = flag ? 9 : bit[pos];
    long long ans = 0;
    for (int i = 0; i <= d; i++)
    {
        int tmp = st + i * (pos - x);
        ans += dp(pos - 1, x, tmp, flag || i != d);
    }
    if (flag) f[pos][x][st] = ans;
    return ans;
}
long long solve(long long n)
{
    if (n < 0) return 0;
    int len = 0;
    while (n > 0)
    {
        bit[++len] = n % 10;
        n /= 10;
    }
    long long res = 0;
    for (int i = 1; i <= len; i++)
    {
        res += dp(len, i, 0, 0);
    }
    return res - len + 1;
}
int main()
{
    int t;
    scanf("%d", &t);
    memset(f, -1, sizeof(f));
    while (t--)
    {
        cin >> a >> b;
        cout << solve(b) - solve(a - 1) << endl;
    }
    return 0;
}