Codeforces 930B Game with String

Codeforces930B游戏字符串策略

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本文探讨了Codeforces930B题目中的一种游戏策略，即通过分析特定字符串及其循环移位来确定玩家获胜的概率。文章提供了一个算法解决方案，通过枚举不同字母开头及各种可能的区间长度来找出唯一确定移位的情况。

传送门：Codeforces 930B

Vasya and Kolya play a game with a string, using the following rules. Initially, Kolya creates a string s, consisting of small English letters, and uniformly at random chooses an integer k from a segment [0, len(s) - 1]. He tells Vasya this string s, and then shifts it k letters to the left, i. e. creates a new string t = sk + 1sk + 2... sns1s2... sk. Vasya does not know the integer k nor the string t, but he wants to guess the integer k. To do this, he asks Kolya to tell him the first letter of the new string, and then, after he sees it, open one more letter on some position, which Vasya can choose.

Vasya understands, that he can't guarantee that he will win, but he wants to know the probability of winning, if he plays optimally. He wants you to compute this probability.

Note that Vasya wants to know the value of k uniquely, it means, that if there are at least two cyclic shifts of s that fit the information Vasya knowns, Vasya loses. Of course, at any moment of the game Vasya wants to maximize the probability of his win.

Input

The only string contains the string s of length l (3 ≤ l ≤ 5000), consisting of small English letters only.

Output

Print the only number — the answer for the problem. You answer is considered correct, if its absolute or relative error does not exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if $\text{[math]}$

  Examples
 
    input
    
     Copy
    
technocup

    output
   
1.000000000000000

    input
    
     Copy
    
tictictactac

    output
   
0.333333333333333

    input
    
     Copy
    
bbaabaabbb

    output
   
0.100000000000000

Note

In the first example Vasya can always open the second letter after opening the first letter, and the cyclic shift is always determined uniquely.

In the second example if the first opened letter of t is "t" or "c", then Vasya can't guess the shift by opening only one other letter. On the other hand, if the first letter is "i" or "a", then he can open the fourth letter and determine the shift uniquely.

分析：概率题，比赛时候没认真看题，觉得比较难直接跳过了QAQ

枚举不同字母开头，区间长度从0到len-1的所有情况，若只出现了一次则记录下来，累加即可

具体实现见代码

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 5000+5;
char str[maxn*2];
int a[26][26][maxn];
int tot,n;

int main(){
    scanf("%s",str+1);
    n = strlen(str+1);
    for (int i=1; i<=n; i++) str[i+n] = str[i];
    memset(a,0,sizeof(a));
    for (int i=1; i<=n; i++)
        for (int j=i+1; j<i+n; j++) a[str[i]-'a'][str[j]-'a'][j-i+1]++;

    tot = 0;
    for (int i=0; i<26; i++) {
        int tmp = 0;
        for (int j=1; j<=n; j++) {
            int sum = 0;
            for (int k=0; k<26; k++) if (a[i][k][j]==1) sum++;
            tmp = max(sum,tmp);
        }
        tot += tmp;
    }
    printf("%.14f\n",double(tot)/n);
    return 0;
}