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

最新推荐文章于 2020-03-08 19:50:52 发布

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最新推荐文章于 2020-03-08 19:50:52 发布

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分类专栏：数位DP

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本文介绍了一种使用数位DP算法解决平衡数问题的方法。平衡数是指在十进制表示下，每个偶数位出现次数为奇数，每个奇数位出现次数为偶数的正整数。文章详细阐述了算法思路，包括如何压缩数字状态和实现递归搜索，最终通过实例演示了如何求解给定区间内的平衡数数量。

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

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

 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：求给定区间[l,r]中奇数的个数是偶数个，偶数的个数是奇数个的数的个数。

思路：数位DP，一共有10个数字0~9，每个数字有三种状态——0：该数字没有出现过，1：该数字出现奇数次，2：该数字出现偶数次。将一个数按这种方案压缩，可以压缩至3^10<60000以内，我们用二维数组dp[20][60000]进行处理即可。

#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
int dit[20];
ll dp[20][60000];
ll judge(int state)//将压缩的十进制数转换为三进制，判断0~9每个数是否符合要求
{
	int a[10];
	for(int i=0;i<10;i++)
	{
		a[i]=state%3;
		state/=3;
	}
	for(int i=0;i<10;i++)
	{
		if(a[i]==0) continue;
		if(i%2==0&&a[i]==2||i%2==1&&a[i]==1)
			return 0;
	}
	return 1;
}
ll change(int pos,int state)//将压缩的十进制数转换为三进制，并改变pos位上的数的奇偶
{
	int a[10],ans=0;
	for(int i=0;i<10;i++)
	{
		a[i]=state%3;
		state/=3;
	}
	if(a[pos]==0) a[pos]=1;
	else if(a[pos]==1) a[pos]=2;
	else if(a[pos]==2) a[pos]=1;
	for(int i=9;i>=0;i--)
	{
		ans=ans*3+a[i];
	}
	return ans;
}
ll dfs(int pos,int state,int lead,int limit)
{
	if(pos<0) return judge(state);
	if(!limit&&!lead&&dp[pos][state]!=-1) return dp[pos][state];
	ll ans=0;
	int up=limit?dit[pos]:9;
	for(int i=0;i<=up;i++)
	{
		ans+=dfs(pos-1,(lead&&i==0)?0:change(i,state),lead&&i==0,limit&&i==dit[pos]);
	}
	if(!limit&&!lead) dp[pos][state]=ans;
	return ans;
}
ll solve(ll x)
{
	int len=0;
	while(x)
	{
		dit[len++]=x%10;
		x/=10;
	}
	return dfs(len-1,0,1,1);
}
int main()
{
	int t;
	ll n,m;
	cin>>t;
	memset(dp,-1,sizeof(dp));
	while(t--)
	{
		cin>>n>>m;
		cout<<solve(m)-solve(n-1)<<endl;
	}
	return 0;
}