HDU 3709 Balanced Number 数位dp

F - Balanced Number

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

Submit  Status  Practice  HDU 3709

Appoint description:  System Crawler  (2016-04-24)

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

      
      
      
      

       
       
       
       2
0 9
7604 24324 
      
      
      
      

 

Sample Output

      
      
      
      

       
       
       
       10
897 
      
      
      
      

 找平衡数

定义平衡数 以数中的一个为中心然后两边的力矩要相等

想了半天没有很好的思路

看了网上的代码发现只要枚举每一位数然后以这个数按位dp就行了

dp[i,j,pre]表示i位数以j为中心的时候左边的比重为pre

注意最后要排除0 ，00， 000，0000，00000，重复的情况所以答案要ans-len+1

ACcode：

#include <cstdio>
#include <cstring>
#define ll long long
ll dp[20][20][1386];
int data[20];
ll dfs(int len,int pos,int pre,bool limit){
    if(!len) return pre==0;
    if(pre<0)return 0;
    if(!limit&&dp[len][pos][pre]!=-1)return dp[len][pos][pre];
    int ed=limit?data[len]:9;
    ll ans=0;
    for(int i=0;i<=ed;++i)
        ans+=dfs(len-1,pos,pre+i*(len-pos),limit&&i==ed);
    return limit?ans:dp[len][pos][pre]=ans;
}
ll fun(ll n){
    int len=0;
    while(n){
        data[++len]=n%10;
        n/=10;
    }
    ll ans=0;
    for(int i=1;i<=len;++i)
        ans+=dfs(len,i,0,1);
    return ans-len+1;
}
void doit(){
    ll x,y;
    scanf("%I64d%I64d",&x,&y);
    printf("%I64d\n",fun(y)-fun(x-1));
}
int main(){
    int loop;memset(dp,-1,sizeof(dp));
    scanf("%d",&loop);
    while(loop--)doit();
}