Hdu-3709 Balanced Number(数位DP)

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本文介绍了一种算法，用于计算给定范围内平衡数的数量。平衡数是指一个非负整数，通过在其某个位上放置支点，使得左侧数字序列与右侧数字序列经过特定计算后的力矩相等。

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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 ≤ 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：问你l到r中有多少个平衡数字，及选取一个轴后左右两边分别乘上1,2,3,4..后相等的数字.



分析:每个数字最多一个轴，枚举所有可能，然后相加，注意特殊处理0(会被多算).


#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<ctime>  
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
#define INF 0x3f3f3f3f
#define eps 1e-9
#define MAXN 100000
using namespace std;
typedef long long ll;
int T;
ll l,r,a[22],dp[22][4000];
ll dfs(int pos,int shaft,int sta,int limit)
{
	if(pos == -1) return sta == 0;
	if(!limit && dp[pos][sta+2000] >= 0) return dp[pos][sta+2000];
	int up = limit? a[pos] : 9;
	ll ans = 0;
	for(int i = 0;i <= up;i++)
	{
		ans += dfs(pos-1,shaft,sta+(pos-shaft)*i,limit && i == up);			
	}  
	if(!limit) dp[pos][sta+2000] = ans;
	return ans;
}
ll solve(ll x)
{
	if(x < 0) return 0;
	int pos = 0;
	while(x)
	{
		a[pos++] = x % 10;
		x /= 10;
	}
	if(!pos) return 1ll; //很隐蔽的错误，x = 0时特殊处理下. 
	ll ans = 0;
	for(int i = 0;i < pos;i++) 
	{
		memset(dp,-1,sizeof(dp));
		ans += dfs(pos-1,i,0,1);
	}
	return ans-pos+1;
}
int main()
{
	scanf("%d",&T);
	while(T--)
	{
		scanf("%I64d%I64d",&l,&r); 
		printf("%I64d\n",solve(r)-solve(l-1));
	}
}