SPOJ - BALNUM - Balanced Numbers（数位DP）

链接：

https://vjudge.net/problem/SPOJ-BALNUM

题意：

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.

思路:

三进制记录每个值用的奇数次还是偶数次。

直接DP即可。

代码：

// #include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<vector>
#include<string.h>
#include<set>
#include<queue>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int MOD = 1e9+7;
const int MAXN = 1e6+10;

ULL a, b;
ULL F[21][60000];
int dig[21];
ULL m[11];

int Upd(int x, int p)
{
    int sum = 0;
    for (int i = 0;i < 10;i++)
    {
        int tmp = x%3;
        x /= 3;
        if (i == p)
            sum += (tmp == 1 ? 2 : 1) * m[i];
        else
            sum += tmp * m[i];
    }
    return sum;
}

bool Check(int x)
{
    int p = 0;
    while(x)
    {
        if (x%3 == 2 && p%2 == 0)
            return false;
        if (x%3 == 1 && p%2 == 1)
            return false;
        x /= 3;
        p++;
    }
    return true;
}

ULL Dfs(int pos, int sta, bool zer, bool lim)
{
    if (pos == -1)
        return Check(sta);
    if (!lim && F[pos][sta] != -1)
        return F[pos][sta];
    int up = lim ? dig[pos] : 9;
    ULL ans = 0;
    for (int i = 0;i <= up;i++)
    {
        ans += Dfs(pos-1, (zer && i == 0) ? 0 : Upd(sta, i), zer && i == 0, lim && i == up);
    }
    if (!lim)
        F[pos][sta] = ans;
    return ans;
}

ULL Solve(ULL x)
{
    int p = 0;
    while(x)
    {
        dig[p++] = x%10;
        x /= 10;
    }
    return Dfs(p-1, 0, 1, 1);
}

int main()
{
    // freopen("test.in", "r", stdin);
    m[0] = 1;
    for (int i = 1;i < 11;i++)
        m[i] = m[i-1]*3;
    memset(F, -1, sizeof(F));
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%llu %llu", &a, &b);
        printf("%llu\n", Solve(b)-Solve(a-1));
    }

    return 0;
}