Palindromic Password

最新推荐文章于 2021-08-18 19:29:53 发布

2020/3/16

最新推荐文章于 2021-08-18 19:29:53 发布

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针对学校IT部门新的密码政策，需要生成由多个6位回文数组成的密码。本篇介绍了一个算法，该算法接收一个6位数并返回与其最接近的回文数，用于帮助自动创建符合要求的密码。

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问题 J: Palindromic Password

时间限制: 3 Sec 内存限制: 128 MB
提交: 212 解决: 60
[ 提交][ 状态][ 讨论版][命题人: admin]

题目描述

The IT department at your school decided to change their password policy. Each password will have to consist of N 6-digit numbers separated by dashes, where N will be determined by the phase of the moon and the weather forecast for the day after it will be generated.
You realized that, if all of the numbers were palindromes (same numbers as the original ones if read backwards), you would have to remember a bunch of 3-digit numbers, which did not sound that bad (at the time).
In order to generate your password of N numbers, you get a list of N randomly generated 6-digit numbers and find the palindromic number closest to them.
Of course, you would like to automate this process...

输入

The first line of the input contains a single positive integer N≤1000 indicating the number of six-digit numbers in the input. Each of the next N lines contains a six-digit number without leading zeroes.

输出

For each six-digit number in the input, output another six-digit number that is closest to it and is also a palindrome. “Closest” in this context means “a number having the smallest absolute difference with the original number”. If there are two different numbers satisfying the above condition, output the smaller one
of the two. Remember, no leading zeroes.

样例输入

2
123321
123322

样例输出

123321
123321

题意：给你一个六位的数字，让你找到一个离它最近的一个回文串，若有两个，输出较小的一个

思路：根据给定数字的前三位，确定一个回文串，然后在这个回文串和原数的差的范围内查找是否有离原数更近的回文串

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;
int main()
{
    int t,a,b,c,aa,p,j,i,flag;
    char s[10];
    scanf("%d",&t);
    while(t--)
    {
        flag=0;
        scanf("%d",&a);
        aa=a/1000;
        b=a/1000;
        while(aa)
        {
            b=b*10+aa%10;
            aa/=10;
        }
        c=fabs(b-a);
        if(b>a)
        {
            for(i=1;i<=c;i++)
            {
                aa=a-i;
                p=aa;
                j=0;
                while(p)
                {
                    s[j++]=p%10-'0';
                    p/=10;
                }
                if(s[0]==s[5]&&s[1]==s[4]&&s[2]==s[3])
                {
                    printf("%d\n",aa);
                    flag=1;
                    break;
                }
            }
        }
        else
        {
            for(i=1;i<=c;i++)
            {
                aa=a+i;
                p=aa;
                j=0;
                while(p)
                {
                    s[j++]=p%10-'0';
                    p/=10;
                }
                if(s[0]==s[5]&&s[1]==s[4]&&s[2]==s[3])
                {
                    flag=1;
                    if(i<c)
                        printf("%d\n",aa);
                    else
                        printf("%d\n",b);
                    break;
                }
            }
        }
        if(flag==0)
        {
            printf("%d\n",b);
        }
    }
    return 0;
}