码迷,mamicode.com
首页 > 其他好文 > 详细

SPOJ - MYQ10 Mirror Number (数位DP)

时间:2014-07-27 11:10:22      阅读:204      评论:0      收藏:0      [点我收藏+]

标签:

Description

A number is called a Mirror number if on lateral inversion, it gives the same number i.e it looks the same in a mirror. For example 101 is a mirror number while 100 is not. 

Given two numbers a and b, find the number of mirror numbers in between them (inclusive of a and b).

Input

First line contains T, number of testcases <= 10^5.
Each testcase is described in a single line containing two numbers a and b.

0 <= a<=b <= 10^44

Output

For each test case print the number of mirror numbers between a and b in a single line.

Example

Input: 3
0 10
10 20
1 4 Output: 3
1
1

题意:求镜像回文的个数在[a,b]区间里

思路:相对较难的数位DP,相对于普通的数位DP ,多了两个参数一个是开始的位置,一个是判断是不是镜像回文,

用一个tmp来储存已经放的数字来判断是不是回文

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
using namespace std;
typedef long long ll;
const int maxn = 50;

ll dp[maxn][maxn][2];
char num[maxn];
int tmp[maxn];

//cur:当前位数,start:镜像回文判断的开始地方,flag:是否是镜像回文,limit:边界判断
ll dfs(int cur, int start, bool flag, bool limit) {
	if (cur == -1)
		return flag;
	if (!limit && dp[cur][start][flag] != -1)
		return dp[cur][start][flag];
	ll ans = 0;
	int end = limit?(num[cur]-48):9;
	for (int i = 0; i <= end; i++) {
		if (i == 0 || i == 1 || i == 8) {
			bool st = (cur==start && i == 0); //判断是否是0情况
			bool newFlag = flag;
			if (flag) {
				if (!st && cur < (start+1)/2)
					newFlag = (tmp[start-cur] == i);
			}
			tmp[cur] = i;
			ans += dfs(cur-1, st?start-1:start, newFlag, limit&&(i==end));
		}
	}
	if (!limit)
		dp[cur][start][flag] = ans;
	return ans;
}

ll cal(char str[])  {
	int len = strlen(str);	
	for (int i = 0; i < len; i++)
		num[i] = str[len-1-i];
	num[len] = 0;
	return dfs(len-1, len-1, 1, 1);
}

int main() {
	int t;
	char a[maxn], b[maxn];
	scanf("%d", &t);
	getchar();
	memset(dp, -1, sizeof(dp));
	while (t--) {
		scanf("%s %s", a, b);
		ll ans = cal(b) - cal(a);
		int len = strlen(a);
		bool flag = true;
		for (int i = 0; i < len; i++)
			if ((a[i] != '0' && a[i] != '1' && a[i] != '8') || (a[i] != a[len-1-i])) {
				flag = false;
				break;
			}
		if (flag)
			ans++;
		printf("%lld\n", ans);
	}
	return 0;
}



SPOJ - MYQ10 Mirror Number (数位DP)

标签:

原文地址:http://blog.csdn.net/u011345136/article/details/38168575

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!