Etaoin Shrdlu 2010.3.2

Etaoin Shrdlu 2010.3.2

Etaoin Shrdlu

Time Limit:1000MS Memory Limit:65536K

Total Submit:27 Accepted:10

Description

The relative frequency of characters innatural language texts is very important for cryptography. However, thestatistics vary for different languages. Here are the top 9 characters sortedby their relative frequencies for several common languages:

English: ETAOINSHR

German: ENIRSATUD

French: EAISTNRUL

Spanish: EAOSNRILD

Italian: EAIONLRTS

Finnish: AITNESLOK

Just as important as the relativefrequencies of single characters are those of pairs of characters, so calleddigrams. Given several text samples, calculate the digrams with the toprelative frequencies.

Input

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

Output

For each test case generate 5 linescontaining the top 5 digrams together with their absolute and relativefrequencies. Output the latter rounded to a precision of 6 decimal places. Iftwo 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 MaxSATand apply the PCP-Theorem [3]

LetF be a set of clauses. The following PTAS algorithm gives an optimal

assignment for F:

0

Sample Output

a 30.073171

!! 3 0.073171

a 30.073171

t 20.048780

oo 2 0.048780

a 80.037209

or 7 0.032558

. 50.023256

e 50.023256

al 4 0.018605

Source

UML 2001

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

#define MAXM 70000
#define MAXN 80+10

struct node
{
int num;
char ch[2];
}hash[MAXM];

int ls,n,tot;
char s[MAXN];

bool cmp(struct node a,struct node b)
{
return (a.num>b.num);
}

void clear()
{
int i;
for(i=0;i<MAXM;i++)
{
hash[i].num=0;
}
}

int main()
{
char mid[20],last;
int i,j,p,t,key;
struct node temp;
while (scanf("%d",&n),n)
{
gets(mid);
tot=0;
clear();
for(i=1;i<=n;i++)
{
gets(s);
ls=strlen(s);
tot+=ls;
for(j=0;j<ls-1;j++)
{
p=256*s[j]+s[j+1];
hash[p].num++;
hash[p].ch[0]=s[j];
hash[p].ch[1]=s[j+1];
}
if (i!=1)
{
p=256*last+s[0];
hash[p].num++;
hash[p].ch[0]=last;
hash[p].ch[1]=s[0];
}
last=s[ls-1];
}
sort(hash,hash+MAXM-1,cmp);
key=hash[4].num;
t=key;
while(hash[t+1].num==key)
t++;
for(i=0;i<=t;i++)
for(j=0;j<i;j++)
if ((hash[i].num==hash[j].num)&&(strcmp(hash[j].ch,hash[i].ch)>0))
{
temp=hash[i];
hash[i]=hash[j];
hash[j]=temp;
}

for(i=0;i<5;i++)
printf("%s %d %.6lf\n",hash[i].ch,hash[i].num,(double)hash[i].num/(double)(tot-1));
printf("\n");
}
return 0;
}