hdu3709 Balanced Number （数位dp）

题目传送门

Balanced Number

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 7994    Accepted Submission(s): 3816

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

 

 

Author

GAO, Yuan

 

 

Source

2010 Asia Chengdu Regional Contest

 

 

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题意：求平衡数，就是以一个数位为支点，左右两边的力矩相等

        比如4139 以3为支点，则4*2+1+1==9*1

题解：数位dp，dp[i][j][k]：i表示数位，j表示支点，k表示力矩和

代码：

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
using namespace std;
typedef long long ll;
ll n,m;
ll dp[20][20][2000];//多加一维表示力矩
int bit[20];
ll dfs(int pos,int center,int sum,bool limit)
{
    if(pos==-1) return sum==0;
    if(sum<0)return 0;//剪枝
    if(!limit&&dp[pos][center][sum]!=-1) return dp[pos][center][sum];
    int up=limit?bit[pos]:9;
    ll ans=0;
    for(int i=0;i<=up;i++)
        ans+=dfs(pos-1,center,sum+(pos-center)*i,limit&&i==up);
    if(!limit) dp[pos][center][sum]=ans;
    return ans;
}
ll calc(ll n)
{
    int len=0;
    while(n)
    {
        bit[len++]=n%10;
        n/=10;
    }
    ll ans=0;
    for(int i=len-1;i>=0;i--)
        ans+=dfs(len-1,i,0,true);
    return ans-(len-1);//排除0、00、000....的干扰
}
int main()
{
    int T;
    scanf("%d",&T);
    memset(dp,-1,sizeof(dp));
    while(T--)
    {
        scanf("%lld %lld",&n,&m);
        printf("%lld\n",calc(m)-calc(n-1));
    }
    return 0;
}