杭电1002 Etaoin Shrdlu

Problem Description

The relative frequency of characters in natural language texts is very important for cryptography. However, the statistics vary for different languages. Here are the top 9 characters sorted by their relative frequencies for several common languages:

English: ETAOINSHR
German:  ENIRSATUD
French:  EAISTNRUL
Spanish: EAOSNRILD
Italian: EAIONLRTS
Finnish: AITNESLOK

Just as important as the relative frequencies of single characters are those of pairs of characters, so called digrams. Given several text samples, calculate the digrams with the top relative frequencies.

Input

 

The input contains several test cases. Each starts with a number n on a separate line, denoting the number of lines of the test case. The input is terminated by n=0. Otherwise, 1<=n<=64, and there follow n lines, each with a maximal length of 80 characters. The concatenation of these n lines, where the end-of-line characters are omitted, gives the text sample you have to examine. The text sample will contain printable ASCII characters only.

 

Output

For each test case generate 5 lines containing the top 5 digrams together with their absolute and relative frequencies. Output the latter rounded to a precision of 6 decimal places. If two digrams should have the same frequency, sort them in (ASCII) lexicographical order. Output a blank line after each test case.

 

Sample Input

2 Take a look at this!! !!siht ta kool a ekaT 5 P=NP Authors: A. Cookie, N. D. Fortune, L. Shalom Abstract: We give a PTAS algorithm for MaxSAT and apply the PCP-Theorem [3] Let F be a set of clauses. The following PTAS algorithm gives an optimal assignment for F: 0

Sample Output

 a 3 0.073171

!! 3 0.073171

a  3 0.073171

 t 2 0.048780

oo 2 0.048780

 

 

 a 8 0.037209

or 7 0.032558

.   5 0.023256

e  5 0.023256

al  4 0.018605

 

题意：输入n行字符串，两个字符为一组；统计他们出现的频率，输出频率的前五位；频率相同按字典序输出；

 

AC代码：

 #include<iostream>
 #include<cstdio>
 #include<cstring>

 using namespace std;

 int main()
 {
     int dp[][]={};//辅助数组。
     char m[];//对字符串的输入
     string str;//合成字符的存储；
     int max=;//存储最大的次数
     int n;//存储行数；
     int j,i;//循环变量；
     int leng;//存储总长度；
     char a,b;
     while(cin>>n)
     {
         gets(m);//录入n后的回车；
         if(n==)break;//当n等于0是跳出；
         memset(dp,,sizeof(dp));
         gets(m);//录入第一行；
         str=m;//存入str中
         for(i=;i<=n;i++)
         {
             gets(m);//录入第i行；
             str+=m;//存入str中；
         }
         leng=str.length();//读出str的长度；
         for(i=;i<=leng-;i++)//逐个检测；
         {
             dp[str[i]][str[i+]]++;
         }
         max=;
         double sum=;
         for(i=;i<;i++)
         {
             for(j=;j<;j++)
             {
                 if(dp[i][j]>max)max=dp[i][j];
                 sum+=dp[i][j];
             }
         }
         n=;
         while(max)
         {
             for(i=;i<;i++)
             {
                 for(j=;j<;j++)
                 {
                     if(dp[i][j]==max)
                     {
                         a=i;
                         b=j;
                         cout<<a<<b<<" "<<max<<" ";
                         printf("%.6f\n",max/sum);
                         n++;
                     }
                     if(n==)break;//输出5组停止
                 }
                 if(n==)break;//输出5组停止
             }
             max--;
             if(max==||n==)break;//输出5组停止，输出不够5组停止；
         }
         cout<<endl;//每个样例输出结束后输出一行空格；
     }
     return ;
 }