hdu3709（数位dp）

最新推荐文章于 2022-04-05 09:12:07 发布

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最新推荐文章于 2022-04-05 09:12:07 发布

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分类专栏：数位DP 文章标签： hdu dp

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本文介绍了一种用于计算特定范围内平衡数数量的高效算法。平衡数定义为一个非负整数，该整数可以通过选择一个数字作为支点，使得支点左侧每个数字乘以其到支点的距离之和等于右侧相应值。文章详细解释了算法的设计思路，并提供了完整的C++实现代码。

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地址：http://acm.hdu.edu.cn/showproblem.php?pid=3709

Balanced Number

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)

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

Sample Output

  
   10
897

题意：找平衡数。平衡数指数中存在某一位数为中心，中心左边个数字分别乘以距离得到的数字之和与右边的相等。

思路：以位数、中心的位置、右边减左边的数值这三个性质为三维打表。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define LL __int64  //注意使用int64
LL dp[20][20][1800];
int len1[20];
LL dfs(int len,int zhou,int sum,bool mz)  //len指数位，zhou指中心，sum指右减左，mz指是否达到上限
{
    if(!len) return sum?0:1;
    if(sum<0) return 0;  //如果sum<0，说明再往后也不行
    if(!mz&&dp[len][zhou][sum]!=-1) return dp[len][zhou][sum];
    int maxx=mz?len1[len]:9;
    LL ans=0;
    for(int i=0;i<=maxx;i++)
        ans+=dfs(len-1,zhou,sum+i*(len-zhou),mz&&i==maxx);
    if(!mz) dp[len][zhou][sum]=ans;
    return ans;
}
LL getans(LL m)
{
    if(m<0) return 0;
    LL len=1;
    while(m)
    {
        len1[len++]=m%10;
        m/=10;
    }
    LL ans=0;
    for(int i=len-1;i>0;i--)
        ans+=dfs(len-1,i,0,1);
    return ans-len+2;  //"-len+2"是因为0与00、000···相同
}
int main()
{
    memset(dp,-1,sizeof(dp));
    int t;
    LL l,r;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d%I64d",&l,&r);
        printf("%I64d\n",getans(r)-getans(l-1));  //这里是r与l-1，不是以前打表时的r+1与l
    }

}