hdu3709——Balanced Number（数位dp）

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

2

0 9

7604 24324

Sample Output

10

897

s表示支点，sum表示计算到pos位时的值

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <set>
#include <cmath>
#include <map>
#include <algorithm>
#define INF 0x3f3f3f3f
#define MAXN 10000000005
#define Mod 10001
using namespace std;
int dight[30];
long long dp[20][20][5000];
long long dfs(int pos,int s,bool limit,int sum)
{
if(pos==0)
return sum==0;
if(!limit&&dp[pos][s][sum]!=-1)
return dp[pos][s][sum];
int end;
long long ret=0;
if(limit)
end=dight[pos];
else
end=9;
for(int d=0;d<=end;++d)
{
ret+=dfs(pos-1,s,limit&&d==end,sum+(pos-s)*d);
}
if(!limit)
dp[pos][s][sum]=ret;
return ret;

}
long long solve(long long a)
{
memset(dight,0,sizeof(dight));
int cnt=1;
while(a!=0)
{
dight[cnt++]=a%10;
a/=10;
}
long long ans=0;
for(int i=1;i<cnt;++i)
ans+=dfs(cnt-1,i,1,0);
cnt--;
return ans-cnt+1;
}
int main()
{
memset(dp,-1,sizeof(dp));
int t;
scanf("%d",&t);
while(t--)
{
long long x,y;
scanf("%I64d%I64d",&x,&y);
printf("%I64d\n",solve(y)-solve(x-1));
}
return 0;
}