SPOJ BALNUM - Balanced Numbers

最新推荐文章于 2019-08-11 22:46:01 发布

原创最新推荐文章于 2019-08-11 22:46:01 发布 · 424 阅读

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本文介绍了一种用于找出特定区间内所有平衡数数量的高效算法。平衡数是一种特殊整数，其特征在于每个偶数位出现奇数次，每个奇数位出现偶数次。通过动态规划方法，文章详细解释了如何计算给定范围内平衡数的数量。

Balanced Numbers

Time Limit:123MS Memory Limit:1572864KB 64bit IO Format:%lld & %llu

Description

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

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>

using namespace std;

typedef long long LL;

int a[20], b[20], base[10];
LL dp[20][59050]; // 3^10=59049

LL dfs(int len, int st, bool all0, bool limit)
{
    if(len < 1)    {
        int i=0;
        while(st) {
            if(i&1 && st%3==1)    return 0;
            if(!(i&1) && st%3==2)    return 0;
            
            st /= 3;
            i++;
        }
        return 1;
    }
    
    if(!limit && !all0 && dp[len][st] != -1)    return dp[len][st];
    
    int maxn = limit ? a[len] : 9;
    LL ret = 0;
    int new_st, temp=st;
    for(int i=0; i<=maxn; i++) {
        if(all0&&i==0) {
            new_st = st;
        }
        else {
            int k = temp%3;
            if(k<2)    new_st = st + base[i];
            else    new_st = st - base[i];
        }
        
        ret += dfs(len-1, new_st, all0&&i==0, limit&&i==maxn);

        temp /= 3;
    }    
    if(!limit && !all0)    dp[len][st] = ret;
    return ret;
}

LL f(LL x)
{
    int len=0;
    while(x) {
        a[++len] = x % 10;
        x /= 10;
    }    
    memset(b, 0, sizeof(b));
    return dfs(len, 0, 1, 1);
}

void Init()
{
    base[0]=1;
    for(int i=1; i<=9; i++) {
        base[i] = base[i-1] * 3;
    }
}

int main ()
{
    Init();
    int T;
    LL A, B;
    scanf("%d", &T);
    memset(dp, -1, sizeof(dp));
    while(T--) {
        scanf("%lld%lld", &A, &B);
        printf("%lld\n", f(B)-f(A-1));
    }
        
    return 0;
}