SPOJ：Balanced Numbers（数位dp）

reference：http://blog.csdn.net/lvshubao1314/article/details/43924709

BALNUM - Balanced Numbers

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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

题意：balance number指一个数某数字的奇偶性与该数字出现次数的奇偶性相反，求区间内多少个balance number。

思路：用三进制保存状态，某数位为0代表没有该数字，为1代表出现奇数次，为2代表出现偶数次，其余就很好解决了。

原题地址：http://www.spoj.com/problems/BALNUM/

# include <iostream>
# include <cstring>
# include <cstdio>
# include <cmath>

using namespace std;
typedef long long LL;
LL dp[21][60000];
int a[21];
int judge(int sta)
{
    int arr[10];
    for(int i=0; i<10; ++i)
    {
        arr[i] = sta%3;
        sta /= 3;
    }
    for(int i=0; i<=9; ++i)
    {
        if(arr[i]==0)
            continue;
        if(i&1 && arr[i]==1 || !(i&1)&& arr[i]==2)
            return 0;
    }
    return 1;
}

int get(int i, int sta)
{
    int arr[10], ans=0;
    for(int i=0; i<10; ++i)
    {
        arr[i] = sta%3;
        sta /= 3;
    }
    if(arr[i]==0)
        arr[i] = 1;
    else
        arr[i] = 3-arr[i];
    for(int i=9; i>=0; --i)
        ans = ans*3 + arr[i];
    return ans;
}

LL dfs(int pos, int sta, bool zero, bool limit)
{
    if(pos == -1)
        return judge(sta);
    if(!limit && dp[pos][sta] != -1) return dp[pos][sta];
    int up = limit?a[pos]:9;
    LL ans = 0;
    for(int i=0; i<=up; ++i)
        ans += dfs(pos-1, (zero&&i==0)?0:get(i, sta), zero&&i==0, limit&&i==a[pos]);
    if(!limit)
        dp[pos][sta] = ans;
    return ans;
}

LL solve(LL num)
{
    int k=0;
    while(num)
    {
        a[k++] = num%10;
        num /= 10;
    }
    return dfs(k-1, 0, true, true);
}

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