HDU 3709 数位dp

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].

 

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

20 97604 24324

 

Sample Output

10897

题意：给你一个区间，求区间内所有平衡数。

平衡数是指可以找到一个支点，左边右边的数乘以力矩加起来之和相等。

题解：数位dp，但要注意排除00,000,这样的情况。

#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;typedef long long ll;ll dp[20][20][2000];ll digit[20];ll dfs(ll len,ll fixloc,ll sum,bool fp){if(!len){return sum==0?1:0;}if(sum<0)return 0;if(!fp&&dp[len][fixloc][sum]!=-1)return dp[len][fixloc][sum];ll ret=0,i,fpmax=fp?digit[len]:9;for(i=0;i<=fpmax;i++){ret+=dfs(len-1,fixloc,sum+i*(len-fixloc),fp&&i==fpmax);}if(!fp)dp[len][fixloc][sum]=ret;return ret;}ll solve(ll n){if(n==-1)return 0;if(n==0)return 1;ll len=0;while(n){digit[++len]=n%10;n/=10;}ll ret=0,i;for(i=len;i>=1;i--){ret+=dfs(len,i,0,true);}return ret-len+1;//排除00,000这些情况}int main(){memset(dp,-1,sizeof(dp));ll t,x,y;scanf("%lld",&t);while(t--){scanf("%lld%lld",&x,&y);printf("%lld\n",solve(y)-solve(x-1));}return 0;}