HDU3709：Balanced Number(数位DP+记忆化DFS)

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 ≤ 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：找出区间内平衡数的个数，所谓的平衡数，就是以这个数字的某一位为支点，另外两边的数字大小乘以力矩之和相等，即为平衡数

思路：按位枚举，找出所有可能的状况进行dfs，与POJ3252类似

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int bit[19];
__int64 dp[19][19][2005];
//pos为当前位置
//o为支点
//l为力矩
//work为是否有上限
__int64 dfs(int pos,int o,int l,int work)
{
    if(pos == -1)
        return l == 0;//已经全部组合完了
    if(l<0)//力矩和为负，则后面的必然小于0
        return 0;
    if(!work && dp[pos][o][l]!=-1)//没有上限，且已经被搜索过了
        return dp[pos][o][l];
    __int64 ans = 0;
    int end = work?bit[pos]:9;//有上限就设为上限，否则就设为9
    for(int i=0; i<=end; i++)
    {
        int next = l;
        next += (pos-o)*i;//力矩
        ans+=dfs(pos-1,o,next,work&&i==end);
    }
    if(!work)
        dp[pos][o][l] = ans;
    return ans;
}

__int64 solve(__int64 n)
{
    int len = 0;
    while(n)
    {
        bit[len++] = n%10;
        n/=10;
    }
    __int64 ans = 0;
    for(int i = 0; i<len; i++)
    {
        ans+=dfs(len-1,i,0,1);
    }
    return ans-(len-1);//排除掉0,00,000....这些情况
}

int main()
{
    int T;
    __int64 l,r;
    scanf("%d",&T);
    memset(dp,-1,sizeof(dp));
    while(T--)
    {
        scanf("%I64d%I64d",&l,&r);
        printf("%I64d\n",solve(r)-solve(l-1));
    }

    return 0;
}