• UVA


    A palindrome is a word, number, or phrase that reads the same forwards as backwards. For example, the name “anna” is a palindrome. Numbers can also be palindromes (e.g. 151 or 753357). Additionally numbers can of course be ordered in size. The first few palindrome numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, ... The number 10 is not a palindrome (even though you could write it as 010) but a zero as leading digit is not allowed.
    Input The input consists of a series of lines with each line containing one integer value i (1 ≤ i ≤ 2∗109). This integer value i indicates the index of the palindrome number that is to be written to the output, where index 1 stands for the first palindrome number (1), index 2 stands for the second palindrome number (2) and so on. The input is terminated by a line containing ‘0’.
    Output
    For each line of input (except the last one) exactly one line of output containing a single (decimal) integer value is to be produced. For each input value i the i-th palindrome number is to be written to the output.
    Sample Input
    1 12 24 0
    Sample Output
    1 33 151

    先观察一个n位数可以产生多少个回文数(不包括少于n位的数字),除去对称的一边,其可以任意改变的位数t=n/2+n%2,其中第一位数不能为0,则其能生成9e(t-1)个回文数,那么通过这个规则,使用while循环,就可以得知要输出的回文的位数。

    然后就是回文排序的问题,这就是一个简单的数学问题,用题目要求的数据的i减去1到n-1位数能生成的回文数,我们得到余数p,p的意义是在一个t位数上第p小的数字,再以回文的形式输出它的另一半即可。

    另外题目中i的最大值位2e9,所以题目中部分数据要用long long。

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    #include<stack>
    #include<map>
    using namespace std;
    const int MAX=1e5+10;
    #define INF 0x7fffffff
    #define ll long long
    #define FOR(i,n) for(i=1;i<=n;i++)
    #define mst(a) memset(a,0,sizeof(a))
    #define mstn(a,n) memset(a,n,sizeof(a))
    //struct num{int a,b,i;}a[1005];
    //bool cmp(const num &x, const num &y){return x.a>y.a;}
    
    int main()
    {
        int n;    
        while(scanf("%d",&n)&&n)
        {
            ll p=9;
            int t=1,i,s[MAX];
            while(n>p)
            {
                t++;
                n-=p;
                if(t>=3&&t%2==1)
                {
                    p*=10;
                }
            }
            int len=t/2+t%2,k=p/9;
            n+=k-1;    
            for(i=len;i>=1;i--)
            {
                s[i]=n%10;
                n/=10;
            }    
            for(i=1;i<=len;i++) cout<<s[i];
            if (t%2) len--;
            for(i=len;i>=1;i--) cout<<s[i];
            cout<<endl;
        } 
        return 0;
    }
  • 相关阅读:
    leetcode206题实现反转链表(c语言)
    V22017编写C/C++时没有与参数列表匹配的重载函数实例
    3DMAX导出到Unity坐标轴转换问题
    ihandy2019笔记编程真题
    模糊数学中合成算子的计算方法
    点击Button按钮实现页面跳转
    做HTML静态页面时遇到的问题总结
    pip换源
    Python正课146 —— DRF 进阶7 JWT补充、基于权限的角色控制、django缓存
    Python正课145 —— DRF 进阶6 自定制频率、接口文档、JWT
  • 原文地址:https://www.cnblogs.com/qq936584671/p/7125853.html
Copyright © 2020-2023  润新知