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http://www.spoj.com/problems/BALNUM/

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

/**
spoj10606 数位dp （求出现的数字，所有偶数出现奇数次，所有奇数出现偶数次）
解题思路：3进制表示数字0~9的出现情况，0表示没有出现，1表示奇数次，2表示偶数次
*/
#include <string.h>
#include <stdio.h>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
LL dp[20][60000];
int bit[20];
bool check(int s)
{
    int num[10];
    for(int i=0;i<10;i++)
    {
        num[i]=s%3;
        s/=3;
    }
    for(int i=0;i<10;i++)
    {
        if(num[i]!=0)
        {
            if(i%2==0&&num[i]==2)return false;
            if(i%2==1&&num[i]==1)return false;
        }
    }
    return true;
}
int getnews(int x,int s)
{
    int num[10];
    for(int i=0;i<10;i++)
    {
        num[i]=s%3;
        s/=3;
    }
    if(num[x]==0)
        num[x]=1;
    else
        num[x]=3-num[x];
    int news=0;
    for(int i=9;i>=0;i--)
    {
        news*=3;
        news+=num[i];
    }
    return news;
}

LL dfs(int pos,int s,int flag,int z)
{
    if(pos==-1)return check(s);
    if(!flag&&dp[pos][s]!=-1)
        return dp[pos][s];
    LL ans=0;
    int end=flag?bit[pos]:9;
    for(int i=0;i<=end;i++)
    {
        ans+=dfs(pos-1,(z&&i==0)?0:getnews(i,s),flag&&i==end,z&&i==0);
    }
    if(!flag)dp[pos][s]=ans;
    return ans;
}
LL solve(LL n)
{
    int len=0;
    while(n)
    {
        bit[len++]=n%10;
        n/=10;
    }
    return dfs(len-1,0,1,1);
}

int main()
{
    int T;
    memset(dp,-1,sizeof(dp));
    LL a,b;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld%lld",&a,&b);
        printf("%lld\n",solve(b)-solve(a-1));
    }
    return 0;
}

spoj10606 数位dp （求出现的数字，所有偶数出现奇数次，所有奇数出现偶数次）

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原文地址：http://blog.csdn.net/lvshubao1314/article/details/43924709