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最新推荐文章于 2019-07-30 11:06:51 发布

原创最新推荐文章于 2019-07-30 11:06:51 发布 · 718 阅读

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本文介绍了一种基于电话号码的分组算法实现。通过字符映射到数字的方式将电话号码转换为整数，并使用快速排序进行排序。接着，对排序后的电话号码进行分组并输出重复号码及其出现次数。

简单模拟题。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<stdlib.h>
using namespace std;
int cmp(const void *a,const void *b)
{
   return *(int *)a - *(int *)b;
}
void ini(int *map)
{
  for (int i = 0; i <= 9; i++)
         map[i + '0'] = i;
     map['A'] = map['B'] = map['C'] = 2;
     map['D'] = map['E'] = map['F'] = 3;
     map['G'] = map['H'] = map['I'] = 4;
     map['J'] = map['K'] = map['L'] = 5;
     map['M'] = map['N'] = map['O'] = 6;
     map['P'] = map['R'] = map['S'] = 7;
     map['T'] = map['U'] = map['V'] = 8;
     map['W'] = map['X'] = map['Y'] = 9;
}

int main()
{
    int map[256];
    char str[256];
    ini(map);
    int n;
    int *s;
    cin >> n;
    s = new int[n];
    for(int i=0; i<n; ++i)
    {
      scanf("%s",str);
      int sum = 0;
      for(int e=0; str[e]!='\0'; ++e)
      {
        if(str[e]=='-')
         continue;
        else
         sum = sum*10 + map[str[e]];
      }
      s[i] = sum;
    }
    qsort(s,n,sizeof(int),cmp);
    int i = 0;
    int flag = 0;
    for(int j=0; j<n; )
    {
      i = j;
      while(s[++i]==s[j]);
      if(i-1!=j)
      {
        printf("%03d-%04d %d\n",s[j]/10000,s[j]%10000,i-j);
       flag = 1;
      }
       j = i;
    }
    if(!flag)
      cout<<"No duplicates."<<endl;
    return 0;
}