Balanced Numbers（数位dp）

Description

Balanced numbers have been used by mathematicians for centuries. A positive integer is considered a balanced number if:

1)      Every even digit appears an odd number of times in its decimal representation

2)      Every odd digit appears an even number of times in its decimal representation

For example, 77, 211, 6222 and 112334445555677 are balanced numbers while 351, 21, and 662 are not.

Given an interval [A, B], your task is to find the amount of balanced numbers in [A, B] where both A and B are included.

Input

The first line contains an integer T representing the number of test cases.

A test case consists of two numbers A and B separated by a single space representing the interval. You may assume that 1 <= A <= B <= 10¹⁹

Output

For each test case, you need to write a number in a single line: the amount of balanced numbers in the corresponding interval

Example

Input:
2
1 1000
1 9

Output:
147
4
开始把题目看错，以为是要求数位中奇数的个数有偶数个，偶数的个数有奇数个
然而也从中找到一个坑点：不要把前导0记录进去，例如009，应该是9,0是没有出现过的
真正的题意是每个奇数出现的次数是偶数，每个偶数出现的次数是奇数。。注意每个。。
如123 我们只能说1是奇数出现了1次，2是偶数出现了1次，3是奇数出现了一次，不能说奇数出现了2次，偶数出现了1次
好吧，其实这道题的难点在于三进制记录状态，因为要在数和状态之间进行互相转换，哈希好像做不了。。

#include <stdio.h>
#include <string.h>
long long dp[][];
int bit[];
int updata(int x,int n)
{
    int num[];
    for(int i=;i<=;i++)
    {
        num[i]=x%;
        x/=;
    }
    if(num[n]==) num[n]=;
    else          num[n]=-num[n];
    int ans=,xx=;
    for(int i=;i<=;i++)
    {
        ans+=num[i]*xx;
        xx*=;
    }
    return ans;
}
int is_ok(int x)
{
    int num[];
    for(int i=;i<=;i++)
    {
        num[i]=x%;
        if(num[i]==&&i%==) return ;
        if(num[i]==&&i%==) return ;
        x/=;
    }
    return ;
}
long long dfs(int len,int x,int flag,int cc)
{
    if(len<=)
    {
        if(is_ok(x)) return ;
        else         return ;
    }
    if(!flag&&dp[len][x]!=-) return dp[len][x];
    long long ans=,tmp;
    int end=flag?bit[len]:;
    for(int i=;i<=end;i++)
    {
        if(cc!=||len==)
        {
           tmp=dfs(len-,updata(x,i),flag&&i==end,);
        }
        else
        {
            if(i==)
            {
               tmp=dfs(len-,x,flag&&i==end,);
            }
            else
            {
                tmp=dfs(len-,updata(x,i),flag&&i==end,);
            }
        }
        ans+=tmp;
    }
    if(!flag) dp[len][x]=ans;
    return ans;
}
long long cacu(long long n)
{
    int pos=;
    if(n<) bit[++pos]=n;
    else
    while(n)
    {
        bit[++pos]=n%;
        n/=;
    }
    return dfs(pos,,,);
}
int main()
{
    memset(dp,-,sizeof(dp));
    int T;
    for(scanf("%d",&T);T;T--)
    {
        long long A,B;
        scanf("%lld%lld",&A,&B);
 //       printf("%lld  %lld\n",cacu(A-1),cacu(B));
        printf("%lld\n",cacu(B)-cacu(A-));
    }
    return ;
}