hdu 3709 Balanced Number(数位DP，5级)

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本文介绍了一种使用数位DP算法来解决寻找特定范围内平衡数数量的问题。平衡数是指可以通过在其某一位上放置支点使左右两边力矩相等的非负整数。文章详细解释了输入输出格式、示例，并提供了完整的C++实现代码。

Balanced Number

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1270 Accepted Submission(s): 553

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

Recommend

zhengfeng

思路：数位DP的模式，只是要注意全是0 的情况得减去。

#include<cstdio>
#include<cstring>
#include<iostream>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
#define LL __int64
#define ll(x) (1<<x)
using namespace std;
LL dp[40][33][2000];//位数,1,0
int bit[40],pos;
LL L,R;int K;
LL DP(int pp,int ba,int dis,bool big)//位置，平衡点，距离量，
{
  if(pp==0)return dis==0;
  if(big&&dp[pp][ba][dis]!=-1)return dp[pp][ba][dis];
  if(dis<0)return 0;
  LL ret=0;
  int kn=big?9:bit[pp];
  FOR(i,0,kn)
  {
    ret+=DP(pp-1,ba,dis+i*(pp-ba),big||kn!=i);
  }
  if(big)dp[pp][ba][dis]=ret;
  return ret;
}
LL get(LL x)
{ pos=0;
  while(x)
  {
    bit[++pos]=x%10;
    x/=10;
  }
  LL ret=0;
  FOR(i,1,pos)
 ret+=DP(pos,i,0,0);
 return ret-pos+1;//全0 5-999 有0 00 000 只能算一种 
}
int main()
{ int cas;
  while(~scanf("%d",&cas))
  {
    FOR(ca,1,cas)
    {clr(dp,-1);
     // cin>>L>>R>>K;
      scanf("%I64d%I64d",&L,&R);
      printf("%I64d\n",get(R)-get(L-1));
      //cout<<get(R)-get(L-1)<<endl;
    }
  }
}