Balanced Number

Balanced Number

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 5199    Accepted Submission(s): 2493

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 ≤ 10¹⁸).

 

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
题目大意就是求给定区间内的平衡数的个数
要明白一点：对于一个给定的数，假设其位数为n，那么可以有n个不同的位作为支点，但每次只能有一个支点
定义dp[len][pos][k]，len表示当前还需处理的位数，pos表示当前的所选的支点的位置，k表示计算到当前的力矩之和（即从最高位到第len+1位
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MOD = 1000000007;
const int maxn = 200010;
int  num[30],len;
LL dp[30][2500][30];
int a[50];
LL dfs(int pos,int zhi,int yi,int last)
{
    if(pos<0) return zhi==0;// 确定完 最高位...最低位 后，如果当前状态合法（zhi == 0），则对应一种合法方案
    if(zhi<0) return 0;//// 如果当前 zhi < 0，那么当前状态肯定不是一个合法的状态，并且之后也不会变为合法状态
    if(!last&&dp[pos][zhi][yi]!=-1)
    {
        return dp[pos][zhi][yi];
    }
    int len=last?num[pos]:9;
    LL res = 0;
    for(int i=0; i<=len; i++)
    {
        res += dfs(pos-1,(zhi+(pos-yi)*i),yi,last&&(i==len));
    }
    if(!last)dp[pos][zhi][yi]= res;
    return res;
}
LL solve(LL n)
{
    len = 0;
   LL sum=0;
    while(n)
    {
        num[len++] = n%10;
        n /= 10;
    }
    for(int i=0;i<len;i++)// 枚举每一个 yi 所在位置
    {
         sum+=dfs(len-1,0,i,1);
    }
    return sum-len+1;//除掉全0的情况，00，0000满足条件，但是重复了 
}
int main()
{
    memset(dp,-1,sizeof(dp));
    int _;
    scanf("%d",&_);
    while(_--)
    {
        LL n,m;
        scanf("%I64d%I64d",&m,&n);
        printf("%I64d\n",solve(n)-solve(m-1));
    }
    return 0;
}