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].

InputThe 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).OutputFor 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

题意：一个数，定一个支点，力矩相等为平衡数。
题解：数位dp，f[i][j][k]表示i为，j位为支点，k为当前力矩。
 1 #include<cstring>
 2 #include<cmath>
 3 #include<algorithm>
 4 #include<iostream>
 5 #include<cstdio>
 6 #define ll long long
 7 using namespace std;
 8 
 9 ll f[21][21][2007],l,r;
10 int a[21],n,cnt;
11 
12 ll dfs(int wei,int zd,int lj,bool flag)
13 {
14     if (wei==0) return lj==0;
15     if (lj<0) return 0;
16     if (!flag&&f[wei][zd][lj]!=-1) return f[wei][zd][lj];
17     ll res=0;
18     int ed;
19     if (flag) ed=a[wei];
20     else ed=9;
21     for (int i=0;i<=ed;i++)
22         res+=dfs(wei-1,zd,lj+(wei-zd)*i,flag&&i==ed);
23     if (!flag) f[wei][zd][lj]=res;
24     return res;    
25 }
26 ll solve(ll x)
27 {
28     cnt=0;
29     while(x)
30     {
31         a[++cnt]=x%10;
32         x/=10;
33     }
34     ll ans=0;
35     for (int i=1;i<=cnt;i++)
36         ans+=dfs(cnt,i,0,1);
37     return ans-(cnt-1);    
38 }
39 int main()
40 {
41     int cas;scanf("%d",&cas);
42     memset(f,-1,sizeof(f));
43     while(cas--)
44     {
45         scanf("%lld%lld",&l,&r);
46         printf("%lld\n",solve(r)-solve(l-1));
47     }
48 }