HDU 3709 Balanced Number-优快云博客

Balanced Number

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2161 Accepted Submission(s): 960

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

Author

GAO, Yuan

Source

2010 Asia Chengdu Regional Contest

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数位DP，可以参考本博客另外两篇博文：

HDU 2089 不要62

HDU 3652 B-number

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<vector>
using namespace std;
typedef long long LL;

//dp[i][j][k],i表示位数,j表示去掉的点的位置,k表示两遍数的计算结果的差
LL dp[20][20][2500];
int num[50];
LL dfs(int pos,int center,int cha,bool limit)
{
    if(pos==0) return cha==0;//结果差为0则返回1
    if(cha<0) return 0;
    if(!limit && dp[pos][center][cha]!=-1) return dp[pos][center][cha];
    int end = limit?num[pos]:9;
    LL ans = 0;
    for(int i = 0;i <= end;i++)
    {
        ans+=dfs(pos-1,center,cha+i*(pos-center),limit&&(i==end));
    }
    if(!limit)
        dp[pos][center][cha] = ans;
    return ans;
}
LL work(LL n)
{
    int len = 0;
    while(n)
    {
        num[++len] = n%10;
        n/=10;
    }
    LL ans = 0;
    for(int i = 1;i <= len;i++)
    {
        ans+=dfs(len,i,0,true);
    }
    return ans-len+1;
}
int main()
{
    int T;
    scanf("%d",&T);
    LL a,b;
    while(T--)
    {
        memset(dp,-1,sizeof(dp));
        scanf("%I64d %I64d",&a,&b);
        printf("%I64d\n",work(b)-work(a-1));
    }
    return 0;
}