【数位DP】 F - Balanced Number HDU - 3709

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本文介绍了一种算法，用于计算在给定范围内平衡数的数量。平衡数是一种特殊的非负整数，当在其某一位设置支点时，左侧部分和右侧部分产生的力矩相等。通过递归深度优先搜索和动态规划的方法，文章详细解释了如何高效地找出特定区间内的所有平衡数。

Balanced Number

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 7771 Accepted Submission(s): 3699

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

0 9

7604 24324

Sample Output

897

Author

GAO, Yuan

Source

2010 Asia Chengdu Regional Contest

#include <bits/stdc++.h>
using namespace std;

int ch[30];
long long dp[30][30][2500];

long long dfs(int zd, int pos, int sum, bool lim)
{
	if (pos == 0)
		return sum == 0;

	if (sum < 0)
		return 0;

	if (!lim && dp[zd][pos][sum] != -1)
		return dp[zd][pos][sum];

	long long res = 0;
	int up = lim ? ch[pos] : 9;
	for (int i = 0; i <= up; i++)
		res += dfs(zd, pos - 1, sum + i * (pos - zd), lim && (i == up));

	if (!lim)
		dp[zd][pos][sum] = res;
	return res;
}

long long solve(long long x)
{
	int w = 0;
	while (x)
	{
		ch[++w] = x % 10;
		x /= 10;
	}

	long long ans = 0;
	for (int i = w; i > 0; i--) /// 枚举支点
		ans += dfs(i, w, 0, 1);

	return ans - w + 1; /// 0 000 重复
}

int main()
{
	int t;
	scanf("%d", &t);
	memset(dp, -1, sizeof dp);
	while (t--)
	{
		long long a, b;
		scanf("%lld %lld", &a, &b);
		printf("%lld\n", solve(b) - solve(a - 1));
	}
	return 0;
}