SPOJ 10606 Balanced Numbers(数位dp)

最新推荐文章于 2019-08-21 21:40:00 发布

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本文介绍了一种使用三进制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，其特点是所有偶数位数字在十进制表示中出现奇数次，所有奇数位数字出现偶数次。通过递归状态转移方程，实现了高效求解。

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

题解：
用三进制表示0~9 出现的状态，0表示没出现过，1出现奇数次，2出现偶数次
代码：

#include<bits/stdc++.h>
using namespace std;
#define ll unsigned long long
ll dp[22][60005];
int a[22];
int getnew(int s,int i)
{
    int t[12];
    for(int i=0;i<10;i++)
    {
        t[i]=s%3;
        s/=3;
    }
    if(t[i]==0)t[i]=1;
    else if(t[i]==1)t[i]=2;
    else t[i]=1;
    s=0;
    for(int i=9;i>=0;i--)
        s=s*3+t[i];
    return s;
}
ll check(int s)
{
    for(int i=0;i<10;i++)
    {
        int t=s%3;
        s/=3;
        if(t)
        {
            if((i&1)&&t==1)return 0;
            else if(i%2==0&&t==2)return 0;
        }
    }
    return 1;
}
ll dfs(int pos,int s,int p,int z)
{
    if(pos==-1)return check(s);
    if(!p&&dp[pos][s]!=-1)
        return dp[pos][s];
    int num=p?a[pos]:9;
    ll ans=0;
    for(int i=0;i<=num;i++)
    {
        ans+=dfs(pos-1,(z&&i==0)?0:getnew(s,i),p&&i==num,z&&i==0);
    }
    if(!p)
        dp[pos][s]=ans;
    return ans;
}
ll cal(ll x)
{
    int pos=0;
    while(x)
    {
        a[pos++]=x%10;
        x/=10;
    }
    return dfs(pos-1,0,1,1);
}
int main()
{
    memset(dp,-1,sizeof(dp));
    int T;scanf("%d",&T);
    while(T--)
    {
        ll x,y;
        scanf("%llu%llu",&x,&y);
        printf("%llu\n",cal(y)-cal(x-1));
    }
    return 0;
}