Codeforces 482C. Game with Strings DP, codeforces482c

Codeforces 482C. Game with Strings DP, codeforces482c

Very good pressure dp questions

D [mask] = x when asked about the mask bit, strings in the x state still cannot be distinguished // It is difficult to pre-process the strings.

Dp [x] indicates the probability of not asking at one time, or asking about status x.

Can see the question of the great god of http://blog.csdn.net/houserabbit/article/details/40658791

C. Game with Stringstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

You play the game with your friend. The description of this game is listed below.

Your friend createsNDistinct strings of the same lengthMAnd tells you all the strings. Then he randomly chooses one of them. He chooses strings equiprobably, I. e. the probability of choosing each ofNStrings equals. You want to guess which string was chosen by your friend.

In order to guess what string your friend has chosen, you are allowed to ask him questions. Each question has the following form: «What character stands on positionPosIn the string you have chosen ?» A string is considered guessed when the answers to the given questions uniquely identify the string. After the string is guessed, you stop asking questions.

You do not have a particle strategy, so as each question you equiprobably ask about a position that hasn't been yet mentioned. your task is to determine the expected number of questions needed to guess the string chosen by your friend.

Input

The first line contains a single integerN(1 digit ≤ DigitNLimit ≤ limit 50)-the number of strings your friend came up.

The nextNLines contain the strings that your friend has created. it is guaranteed that all the strings are distinct and only consist of large and small English letters. besides, the lengths of all strings are the same and are between 1 to 20 inclusive.

Output

Print the single number-the expected value. Your answer will be considered correct if its absolute or relative error doesn't exceed 10 seconds-since 9.

Sample test (s) input

2aabaac

Output

2.000000000000000

Input

3aaAaBaCaa

Output

1.666666666666667

Input

3acavacwqq

Output

1.000000000000000

Note

In the first sample the strings only differ in the character in the third position. So only the following situations are possible:

• You guess the string in one question. The event's probability is;
• You guess the string in two questions. The event's probability is · = (as in this case the first question shocould ask about the position that is other than the third one );
• You guess the string in three questions. The event's probability is... =;

Thus, the expected value is equal

In the second sample we need at most two questions as any pair of questions uniquely identifies the string. So the expected number of questions is.

In the third sample whatever position we ask about in the first question, we immediately identify the string.

/*************************************** * ******** Author: CKbossCreated Time: October Sunday 21 seconds File Name: CF482C_C.cpp *************************************** * ********/# include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <string> # include <cmath> # include <cstdlib> # include <vector> # include <queue> # include <set> # include <map> using namespace std; typedef long int LL; char str [55] [22]; int n, m; LL d [(1 <20) + 10]; double dp [(1 <20) + 10]; int countb (LL x) {int ans = 0; while (x) {ans ++; x = x & (x-1LL);} return ans;} int main () {// freopen ("in.txt", "r", stdin ); // freopen ("out.txt", "w", stdout); scanf ("% d", & n); for (int I = 0; I <n; I ++) scanf ("% s", str [I]); m = strlen (str [0]); for (int I = 0; I <n; I ++) {for (int j = 0; j <n; j ++) {if (j = I) continue; /// string I and string j has public posint same = 0; for (int k = 0; k <m; k ++) {if (str [I] [k] = str [j] [k]) same | = (1 <k );} d [same] | = (1LL <I); d [same] | = (1LL <j );}} // d [mask] <= d [mask ^ (1 <I)] for (int mask = (1 <m)-1; mask --) {for (int j = 0; j <m; j ++) {if (mask> j) & 1) {int nmask = mask ^ (1 <j); d [nmask] | = d [mask] ;}} dp [0] = 1; double ans = 0; for (int mask = 0; mask <(1 <m); mask ++) {int c = countb (mask); for (int j = 0; j <m; j ++) {if (mask> j) & 1) = 0) {int nmask = mask | (1 <j ); dp [nmask] + = dp [mask]/(m-c) ;}}for (int j = 0; j <n; j ++) {if (d [mask]> j) & 1LL) ans + = dp [mask] ;}} printf ("%. 15lf \ n ", ans/n); return 0 ;}