[数位DP + 三进制状态压缩] K - Balanced Numbers SPOJ - BALNUM

最新推荐文章于 2022-07-31 00:45:14 发布

原创最新推荐文章于 2022-07-31 00:45:14 发布 · 355 阅读

2 ·

CC 4.0 BY-SA版权

数位DP 专栏收录该内容

9 篇文章

订阅专栏

本文介绍了一种使用状态压缩动态规划方法来计算指定区间内平衡数的数量。平衡数是一种特殊的整数，其特点在于所有偶数位上的数字出现次数为奇数次，而所有奇数位上的数字出现次数为偶数次。文章详细解释了如何通过三进制状态压缩来高效地解决这一问题。

BALNUM - Balanced Numbers

no tags

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

Submit solution!

/// 每个数出现次数的可能性 0 奇数 偶数
/// 分别用0 1 2 表示 
/// 三进制的状态压缩
#include <bits/stdc++.h>
using namespace std;

int ch[30];
long long dp[30][60000];
int mi[11] = {1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049};

bool judge(int sum)
{
	int a[10];  // 各数字出现次数状态
	for (int i = 9; i >= 0; i--)
	{
		a[i] = sum / mi[i];
		sum %= mi[i];
	}

	bool ff = 1;
	for (int i = 9; i >= 0; i--)
	{
		if (a[i] == 0)
			continue;
		if (i % 2 == 0 && a[i] == 2)
			ff = 0;
		if (i % 2 == 1 && a[i] == 1)
			ff = 0;
	}

	if (!ff)
		return 0;
	else
		return 1;
}

int change (int sum, int i)
{
	int t = (sum % mi[i + 1]) / mi[i]; // i 出现次数的状态
	if (t == 2)
		sum -= mi[i];
	else
		sum += mi[i];
	return sum;
}

long long dfs(int pos, int sum, bool lim)
{
	if (pos == 0)
		return judge(sum);

	if (!lim && dp[pos][sum] != -1)
		return dp[pos][sum];

	long long res = 0;
	int up = lim ? ch[pos] : 9;
	for (int i = 0; i <= up; i++)
	{
		/// 前导零
		res += dfs(pos - 1, sum == 0 && i == 0 ? 0 : change(sum, i), lim && (i == up));
	}
	if (!lim)
		dp[pos][sum] = res;
	return res;
}

long long solve(long long x)
{
	int w = 0;
	while (x)
	{
		ch[++w] = x % 10;
		x /= 10;
	}
	return dfs(w, 0, 1);
}

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