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].
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
题意：找出指定区间的平衡数的个数（平衡数是指以一个位为中心轴，左力矩等于右力矩。）
PS：还是不会确定状态啊 只想到左右相等（结果M了），怎么就想不到和为0呢。
解释一下dfs函数四个参数以及状态的含义。
len：数到第几位
sum：数到当前位的力矩之和（数到最后为0的话就是平衡数）
piv：中心轴所在的位
limit：是否为上限
dp[i][j][k]: i:多少位 j:中心轴所在的位 k:数到当前位的力矩之和
（这里有个玄学，我初值赋0会T,赋-1就没事，搞不懂）
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#define LL long long
using namespace std;
int digit[20];
LL dp[20][20][2000];
LL dfs(int len,int sum,int limit,int piv)
{
if(len==0) return sum==0;
if(sum<0) return 0;
//如果sum小于0，因为是从右向左数，后面的肯定是负数，直接返回0
if(!limit && dp[len][piv][sum]!=-1) return dp[len][piv][sum];
LL ans=0;
int up=limit?digit[len]:9;
for(int i=0;i<=up;i++)
{
ans+=dfs(len-1,sum+i*(len-piv),limit&&i==up,piv);
}
if(!limit) dp[len][piv][sum]=ans;
return ans;
}
LL cal(LL a)
{
int len=0;
while(a)
{
digit[++len]=a%10;
a/=10;
}
LL ans=0;
for(int i=1;i<=len;i++)
{
//memset(dp,0,sizeof(dp));
ans+=dfs(len,0,1,i);
}
return ans-(len-1);
}
int main()
{
int t;
scanf("%d",&t);
memset(dp,-1,sizeof(dp));
while(t--)
{
LL x,y;
scanf("%lld%lld",&x,&y);
printf("%lld
",cal(y)-cal(x-1));
}
return 0;
}