HDU 3709 Balanced Number （数位DP）

                                                                                                          Balanced Number            

Time Limit: 5000MS                            Memory Limit: 65535KB 64bit IO Format:                            %I64d & %I64u                        

Submit                                        Status                                                

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

20 97604 24324

Sample Output

10897

Hint

Source

2010 Asia Chengdu Regional Contest

 

题意：在区间中输出满足条件数的个数（条件：对一个数，如果它左右两边的是平衡的，例4139，以3为支点，左面 4*2+1=9，右面9，左右相等）

题解：数位DP，dp[i][j][k],当前位置为i，支点为j，k为左右两边的和（假设左正右负）。

我们枚举支点即可，一个优化，由于是从左往右的，sum的值不能小于0，

代码：

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>#define ll long longusing namespace std;ll dp[30][30][1650];int a[30];ll dfs(int pos,int p,int sum,bool limit){    if(pos==0)        return sum==0;    if(sum<0)        return 0;    if(!limit&&dp[pos][p][sum]!=-1)        return dp[pos][p][sum];    int up=limit?a[pos]:9;    ll ans=0;    for(int i=0;i<=up;i++)    {        ans+=dfs(pos-1,p,sum+i*(pos-p),limit&&i==a[pos]);    }    if(!limit)        dp[pos][p][sum]=ans;    return ans;}ll solo(ll x){    if(x==-1)        return 0;        int pos=0;    while(x)    {        a[++pos]=x%10;        x/=10;    }    ll ans=0;    for(int i=1;i<=pos;i++)        ans+=dfs(pos,i,0,true);    return ans-pos+1;}int main(){    int t;    cin>>t;    memset(dp,-1,sizeof(dp));    while(t--)    {        ll l,r;        cin>>l>>r;        cout<<solo(r)-solo(l-1)<<endl;    }}