SPOJ BALNUM BALNUM - Balanced Numbers

最新推荐文章于 2019-09-26 17:20:24 发布

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本文介绍了一种使用数位DP和三进制状态压缩的方法来计算特定区间内平衡数的数量。平衡数是指其十进制表示中每个奇数位出现偶数次，每个偶数位出现奇数次的正整数。

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
4

题意：求区间奇数出现偶数次，偶数出现奇数次的数有多少。

思路：数位dp+三进制状态压缩，0表示没出现过，1表示出现奇数次，2表示出现偶数次，下面给代码：

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<bitset>
#include <utility>
using namespace std;
#define maxn 60000
typedef long long LL;
LL dp[25][maxn];
int digit[maxn][11];
int num[25];
int three[11] = { 1 };//three求出3的n次方 
bool jud(int status){
	for (int i = 0; i < 10; i++){
		if (i & 1 && digit[status][i + 1] & 1)
			return false;
		else if (!(i & 1) && digit[status][i + 1] == 2)
			return false;
	}
	return true;
}
int getstatus(int status,int now){
	if (digit[status][now + 1] <= 1)
		return status + three[now];
	else 
		return status - three[now];
}
LL dfs(int pos, int status, int limit){
	if (pos < 1){
		return jud(status) ? 1 : 0;
	}	
	if (!limit&&~dp[pos][status])
		return dp[pos][status];
	int end = limit ? num[pos] : 9;
	LL ans = 0;
	for (int i = 0; i <= end; i++){
		if (!i&&!status)
			ans += dfs(pos - 1, 0, limit&&i == end);
		else
			ans += dfs(pos - 1, getstatus(status, i), limit&&i == end);
	}
	if (!limit)
		dp[pos][status] = ans;
	return ans;
}
LL solve(LL x){
	int len = 0;
	while (x){
		num[++len] = x % 10;
		x /= 10;
	}
	return dfs(len, 0,  1);
}
int main(){
	memset(dp, -1, sizeof(dp)); 
	for (int i = 1; i <= 10; i++){
		three[i] = three[i - 1] * 3;
	}
	for (int i = 1; i<three[10]; i++){
		int temp = i;
		for (int j = 1; j<11; j++){
			digit[i][j] = temp % 3;
			temp /= 3;
			if (!temp)
				break;
		}
	}
	int t;
	scanf("%d", &t);
	while(t--){
		LL l, r;
		scanf("%lld%lld", &l, &r);
		printf("%lld\n", solve(r) - solve(l - 1));
	}
}