HDOJ-三部曲-1002-Etaoin Shrdlu

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Etaoin Shrdlu

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 59   Accepted Submission(s) : 12

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

 

 

 

 

这题我想了个新方法，以前那个超时的办法就删了。我以各个字符的ASCII码作为数组下标，建立一个二维数组，来存贮各个双字符组合的数量，最后找出数量最大的那五个。

 

 

#include<iostream>
#include<string.h>
#include<iomanip>
#include<stdio.h>
using namespace std;
struct digram
{
	char c1,c2;
	int num;
}dig[5];                                              //用来存储符合条件的5个双字符组合
int main()
{
	int n,i,j;
	char s[64][81];
	while(cin>>n&&n)
	{
		getchar();
		int ascii[128][128]={0},k=0;
		char let[10000];
		for(i=0;i<n;i++)
		{
			cin.getline(s[i],80);
			int size=strlen(s[i]);
			for(j=0;j<size;j++)
				let[k++]=s[i][j];      //存储各个字符
		}
		int total=k-1;                         //双字符总数
		for(i=0;i<k-1;i++)                     //统计各种双字符组合的个数
			ascii[let[i]][let[i+1]]++;
		for(i=0;i<5;i++)                       //寻找符合条件的5个双字符组合
		{
			dig[i].num=0;
			for(j=0;j<128;j++)
			{
				for(k=0;k<128;k++)
					if(dig[i].num<ascii[j][k]||dig[i].num==ascii[j][k]&&(dig[i].c1>j||dig[i].c1==j&&dig[i].c2>k))
					{
						dig[i].num=ascii[j][k];
						dig[i].c1=j;
						dig[i].c2=k;
					}
			}
			ascii[dig[i].c1][dig[i].c2]=0;
		}
		for(i=0;i<5;i++)
			cout<<dig[i].c1<<dig[i].c2<<' '<<dig[i].num<<' '<<setiosflags(ios::fixed)<<setprecision(6)<<1.0*dig[i].num/total<<endl;
		cout<<endl;
	}
}