HDU 3709 Balanced Number(数位dp)

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本文介绍了一种计算特定范围内平衡数数量的算法。平衡数是指一个非负整数，通过在其某一位上放置支点，使得左侧部分和右侧部分产生的力矩相等。文章详细解释了如何枚举每一位作为中心来判断是否满足平衡条件，并提供了实现这一过程的C++代码。

Balanced Number

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 7215 Accepted Submission(s): 3464

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

题解：
枚举以每一位为中心，中心左边加，右边减，最后若是0则满足条件

代码：
#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll dp[22][22][5005];
int a[22];
ll dfs(int pos,int id,int s,bool p)
{
    if(pos==-1)return s==0;
    if(s<0)return 0;
    if(!p&&dp[pos][id][s]!=-1)
        return dp[pos][id][s];
    int num=p?a[pos]:9;
    ll ans=0;
    for(int i=0;i<=num;i++)
    {
        ans+=dfs(pos-1,id,s+(pos-id)*i,p&&(i==num));
    }
    if(!p)
        dp[pos][id][s]=ans;
    return ans;
}
ll cal(ll x)
{
    int pos=0;
    while(x)
    {
        a[pos++]=x%10;
        x/=10;
    }
    ll ans=0;
    for(int i=0;i<pos;i++)
        ans+=dfs(pos-1,i,0,1);
    return ans-pos+1;
}
int main()
{
    memset(dp,-1,sizeof(dp));
    ll x,y;int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld%lld",&x,&y);
        printf("%lld\n",cal(y)-cal(x-1));
    }
    return 0;
}