微软2014年在线笔试题前两题

本文介绍了一个简单的算法,该算法能够接收包含特定ASCII字符('0'-'9' 和 'a'-'z')的字符串,并按照ASCII顺序进行排序。排序后的字符被分成多个段落,每个段落内的字符严格递增排列,且后续段落为前一段的子集或完全相同。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Description
  For this question, your program is required to process an input string containing only ASCII characters between ‘0’ and ‘9’, or between ‘a’ and ‘z’ (including ‘0’, ‘9’, ‘a’, ‘z’).
  Your program should reorder and split all input string characters into multiple segments, and output all segments as one concatenated string. The following requirements should also be met,
  1. Characters in each segment should be in strictly increasing order. For ordering, ‘9’ is larger than ‘0’, ‘a’ is larger than ‘9’, and ‘z’ is larger than ‘a’ (basically following ASCII character order).
  2. Characters in the second segment must be the same as or a subset of the first segment; and every following segment must be the same as or a subset of its previous segment.
  Your program should output string “” when the input contains any invalid characters (i.e., outside the ’0′-’9′ and ‘a’-’z’ range).
  Input
  Input consists of multiple cases, one case per line. Each case is one string consisting of ASCII characters.
  Output
  For each case, print exactly one line with the reordered string based on the criteria above.
  样例输入
  aabbccdd
  007799aabbccddeeff113355zz
  1234.89898
  abcdefabcdefabcdefaaaaaaaaaaaaaabbbbbbbddddddee
  样例输出
  abcdabcd
  013579abcdefz013579abcdefz
  
  abcdefabcdefabcdefabdeabdeabdabdabdabdabaaaaaaa
我只实现了要求一,后面的还要再想想

#include <iostream>
#include <string>
using namespace std;

extern int max1=0;

void f(string s,int hash[]){
    int i,t;
    //将输入的字符存到hash数组其ASCII码对应的下标下,同时标记重复字符的最大值
    for(i= 0;i<s.size();i++)
    {
        t=int(s[i]);
        hash[t]++;
        if(hash[t]>max1)
            max1=hash[t];
    }

}

int _tmain(int argc, _TCHAR* argv[])
{
    string s;

    cout<<"Input:"<<endl;
    char c;

    getline(cin,s);
    int hash[200]={};
    f(s,hash);

    //遍历hash数组
    while(max1)
    {
        int i=0;
        for(i;i<200;i++)
        {
            if(hash[i]!=0)
            {
                if(hash[i]==max1)
                {
                    max1--;
                }
            c=(char)i;
            cout<<c;
            hash[i]--;
            }

      }
    }

    getchar();
    getchar();

    return 0;
}

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值