hdu3709 Balanced Number

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

这是一道数位dp,建立一个数组dp[pos][zhidian][liju],pos代表当前位置,zhidian代表取的支点,liju代表到当前位置累加的力矩，这个数组表示这些状态下的总个数，初始化为-1。然后注意有上限，上限的意思是:如果范围是[1,12345]，现在找到123，那么仍然有限制，如果再找2变成1232，那么此时没有限制，因为下一位可以找到9.

<pre name="code" class="cpp">#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
#define inf 99999999
#define pi acos(-1.0)
#define maxn 1000050
#define MOD 1000000007
using namespace std;
typedef long long ll;
typedef long double ldb;
int len;
int wei[22];
ll dp[20][20][2000];

ll dfs(int pos,int o,ll pre,int flag){
    int i,j;
    if(pos==0){
        if(pre==0)return 1;
        return 0;
    }
    if(pre<0)return 0;
    if(flag==0 && dp[pos][o][pre]!=-1){
        return dp[pos][o][pre];
    }
    ll ans=0;
    int ed=flag?wei[pos]:9;
    for(int i=0;i<=ed;i++){
        int next=pre;
        next+=(pos-o)*i;
        ans+=dfs(pos-1,o,next,flag&&i==ed );
    }
    if(!flag)dp[pos][o][pre]=ans;
    return ans;
}


ll solve(ll x)
{
    if(x==-1)return 0;
    if(x==0)return 1;
    int i,j,o;
    ll t=x;
    len=0;
    while(t){
        wei[++len]=t%10;
        t/=10;
    }
    ll ans=0;
    for(o=len;o>=1;o--){
        ans+=dfs(len,o,0,1);
    }
    ans-=(len-1);
    return ans;
}


int main()
{
    ll n,m;
    int i,j,T;
    memset(dp,-1,sizeof(dp));
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld%lld",&n,&m);
        printf("%lld
",solve(m)-solve(n-1));
    }
    return 0;
}