hdu-3709 Balanced Number

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].

 

题目先给出平衡数的概念：数n以数n中的某个位为支点，每个位上的数权值为（数字xi*（posi - 支点的posi）），如果数n里有一个支点使得所有数权值之和为0那么她就是平衡数。比如4139，以3为支点，左边 = 4 * (4 - 2) + 1 * (3  - 2) = 9，右边 = 9 * (1 - 2) = -9，左边加右边为0，所以4139是平衡数。现在给出一个区间[l,r]，问区间内平衡数有多少个？

 

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

 

Recommend

zhengfeng   |   We have carefully selected several similar problems for you:  3711 3715 3718 3713 3712 

 

Solution

数位dp

dp[i][j][k]表示考虑到第i位，j为支点，当前权值和为k的数的个数.

先枚举一个支点，再dp

注意全为0的情况被多算了，要减去

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#define lo long long
using namespace std;
int a[23];
lo dp[23][23][3011];
inline long long read()
{
	register long long ans=0,f=1;char ch=getchar();
	while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
	while(isdigit(ch)) {ans=ans*10+ch-'0';ch=getchar();}
	return ans*f;
}
lo dfs(int wi,int mid,int v,bool lim)
{
	if(wi<1)
	  return v==0;
	if(!lim&&dp[wi][mid][v]>-1)
	  return dp[wi][mid][v];
	int o=lim? a[wi]:9;
	lo ans=0;
	for(int i=0;i<=o;i++)
	  ans+=dfs(wi-1,mid,v+(wi-mid)*i,lim&&i==a[wi]);
	if(!lim)
	  dp[wi][mid][v]=ans;
	return ans;
}
lo sol(lo x)
{
	if(x<0) 
	  return 0;
	int w=0;lo ans=0;
	while(x)
	{
		a[++w]=x%10;
		x/=10;
	}
	for(int i=1;i<=w;i++)
	  ans+=dfs(w,i,0,1);
	return ans-w+1;    ////全是0的情况被多算了 
}
int main()
{
	int t;lo l,r;
	t=read();
	memset(dp,-1,sizeof(dp));
	while(t--)
	{
		l=read();r=read();
		printf("%lld
",sol(r)-sol(l-1));
	}
	return 0;
}