Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

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Background

Please, help the crazy linguist! He is trapped in his madness. He has developed a ``Spanish Beauty Criterion'' for words. It is defined as follows. Given a word w

(where n is the length of the word), he is interested in words which are formed by letters following pattern:

and, also, in his madness, he won't allow a word that actually has i, j, and k, so that (that is, any letter can appear in the word at most two times).

Then, the ``Spanish Beauty Criterion'' (SBC) is defined as:

where P is defined as the probability of appearance of a letter in Spanish, defined by the following table:

So, given a word w, of size n, our poor linguist wants to know if this word is above or below the average of the SBC of all the words of size n that can be constructed following the above pattern and start with the same letter than w.

The Input

The input will have a first line with a number, N, the number of samples that will be entered to know if they are above or below the average. Following this first line, N lines with a word in each one (of at most seven characters each). All the input words follow the given pattern.

The Output

The output will have N lines, each line corresponding to one input word in order, showing just ``above or equal'' or ``below'', depending on the value of the SBC of that word relative to the average of those of the same size.

Sample Input

5
bubu
terabit
hacer
qed
loco

Sample Output

below
above or equal
above or equal
above or equal
above or equal

---

題目描述：
給定一個字串，問這個字串是在 平均 SBC 之上還是之下。
平均 SBC 的取樣：相同字元開頭，且相同長度，產生方式如上描述。

題目解法：
基本上可以本地建表，畢竟每次都要窮舉一次很慢，但是會過就算了。

#include <stdio.h>
#include <string.h>
int count[26], n;
char s[105];
double prob[26] = {
    12.53, 1.42, 4.68, 5.86, 13.68, 0.69, 1.01, 0.70, 6.25, 0.44,
    0.00, 4.97, 3.15, 6.71, 8.68, 2.51, 0.88, 6.87, 7.98, 4.63,
    3.93, 0.90, 0.02, 0.22, 0.90, 0.52
};
char odd[] = {'a', 'e', 'i', 'o', 'u'};
char even[] = {'b', 'c', 'd', 'f', 'g', 'h', 'j', 'k', 'l', 'm', 'n', 'p', 'q', 'r', 's', 't', 'v', 'w', 'x', 'y', 'z'};
double SBC(char s[]) {
    double ret = 0;
    int i;
    for(i = 0; s[i]; i++)
        ret += (i+1)*prob[s[i]-'a'];
    return ret;
}
double avgsbc;
int cnt;
void dfs(int idx) {
    if(idx == n) {
        s[idx] = '\0';
        avgsbc += SBC(s);
        cnt++;
        return;
    }
    int i;
    if(idx&1) {
        for(i = 0; i < 5; i++) {
            if(count[odd[i]-'a'] > 0) {
                count[odd[i]-'a']--;
                s[idx] = odd[i];
                dfs(idx+1);
                count[odd[i]-'a']++;
            }
        }
    } else {
        for(i = 0; i < 21; i++) {
            if(count[even[i]-'a'] > 0) {
                count[even[i]-'a']--;
                s[idx] = even[i];
                dfs(idx+1);
                count[even[i]-'a']++;
            }
        }
    }
}
int main() {
    int testcase;
    int i;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%s", s);
        memset(count, 0, sizeof(count));
        double sbc = SBC(s);
        for(i = 0; s[i]; i++)
            count[s[i]-'a']++;
        for(i = 0; i < 26; i++)
            count[i] = 2;
        count[s[0]-'a'] = 1;
        avgsbc = 0, cnt = 0;
        n = strlen(s);
        dfs(1);
        avgsbc /= cnt;
        puts(sbc < avgsbc ? "below" : "above or equal");
    }
    return 0;
}