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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Problem F: Frequent Substrings

You are working for the R&D department of International Embedded Equipments Enterprise (IEEE). One day, your boss calls you up and tells this: 

"Yes, I think we should use the N-Gram Analysis for the project. What do you think ??"
"Ye...s, I also thought so...", you replied.

After the conversation, having no clue about what he said, you start searching the internet and find the following sentences.

N-Grams are substrings of length N. N-Gram Analysis is a method of matching text, based on the statistical similarity of occurrences of N-Grams. N-Gram Analysis is used in research areas such as speech recognition, database interfacing, network communication etc. 

After going through the topic a few times and reading several articles, you figure out the main parts of the project you have to accomplish. One of the parts required you to do this: Given a string S, you have to find the most frequently occurring N-Gram in S. Since you found this part interesting, you decide to complete it soon. Note that the occurrences can partially overlap. That is, the string bcbcbc has two occurrences of the substring cbc.

Input

Input consists of several test cases. Each test case begins with a line specifying the string S. S will contain no more than 1,000 characters. S can contain any of the printable ASCII characters. Capitalization should be ignored while finding the N-Grams. The next line specifies an integer T. Next T lines give the values for N, 0< N£ Length(S).

Output

For each test case, output T lines specifying the number of occurrences of the N-Gram followed by the N-Gram itself separated by exactly one space. If there are several such N-Grams, output the least lexicographical one, compared in terms of ASCII values.

Note :  Given two strings a=a₀a₁a₂...a_m and b=b₀b₁b₂...b_m over the same alphabet V, we say that a is lexicographically less than b if there exist an integer j, 0£ j£ m, so that a_i=b_i for all i=0,1,...,j-1 and a_j< b_j.

Sample Input

In theory, there is no difference between theory and practice, but in practice, there is.
2
4
9

Sample Output

4  the
2  practice

---

題目要求找到長度為 N 的頻率最高的子字串。

這題很坑爹的是，沒有看到' '(空白字元)出現在 the 跟 practice 前面，一度以為題目有問題。

建造高度數組出來，然後掃描高度數組，最長區間都大於等於 N 的最小索引值。

但要特別處理 N = 1 的時候。


// 0.012s

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct SuffixArray {
    int sa[1005], h[1005], n;
    char str[1005];

    int w[1005], ta[1005], tb[1005]; // buffer
    void sort(int *x, int *y, int m) { // radix sort
        static int i;
        for(i = 0; i < m; i++)
            w[i] = 0;
        for(i = 0; i < n; i++)
            w[x[y[i]]]++;
        for(i = 1; i < m; i++)
            w[i] += w[i-1];
        for(i = n-1; i >= 0; i--)
            sa[--w[x[y[i]]]] = y[i];
    }
    bool cmp(int *r, int a, int b, int l) {
        if(r[a] == r[b]) {
            if(a+l >= n || b+l >= n)
                return false;
            return r[a+l] == r[b+l];
        }
        return false;
    }
    void build_h() {
        int i, j, k;
        for(i = 0; i < n; i++)  ta[sa[i]] = i;
        for(i = 0; i < n; i++) {
            if(ta[i] == 0) {
                h[ta[i]] = 0;
                continue;
            }
            if(i == 0 || h[ta[i-1]] <= 1)
                k = 0;
            else
                k = h[ta[i-1]]-1;
            while(str[sa[ta[i]-1]+k] == str[sa[ta[i]]+k])
                k++;
            h[ta[i]] = k;
        }
    }
    void build() {// x: rank, y: second key(array index)
        int i, k, m = 128, p;
        int *x = ta, *y = tb, *z;
        n = strlen(str);
        x[n] = 0;
        for(i = 0; i < n; i++)
            x[i] = str[i], y[i] = i;
        sort(x, y, m);
        for(k = 1, p = 1; p < n; k *= 2, m = p) {
            for(p = 0, i = n-k; i < n; i++)
                y[p++] = i;
            for(i = 0; i < n; i++) {
                if(sa[i] >= k) {
                    y[p++] = sa[i]-k;
                }
            }
            sort(x, y, m);
            z = x, x = y, y = z;
            for(i = 1, p = 1, x[sa[0]] = 0; i < n; i++)
                x[sa[i]] = cmp(y, sa[i-1], sa[i], k) ? p-1 : p++;
        }
    }
};

int main() {
    SuffixArray SA;
    while(gets(SA.str)) {
        SA.n = strlen(SA.str);
        int T, N, i;
        for(i = 0; i < SA.n; i++) {
            if(SA.str[i] >= 'A' && SA.str[i] <= 'Z')
                SA.str[i] = SA.str[i]-'A'+'a';
        }
        SA.build();
        SA.build_h();
        /*for(i = 0; i < SA.n; i++)
            printf("%s\n%d\n", SA.str+SA.sa[i], SA.h[i]);*/
        scanf("%d", &T);
        while(T--) {
            scanf("%d", &N);
            int mx_f = 0, mx_f_arg = 0;
            for(i = 0; i < SA.n; i++) {
                if(SA.h[i] >= N) {
                    int cnt = 1, j = i;
                    while(j < SA.n && SA.h[j] >= N)
                        cnt++, j++;
                    if(cnt > mx_f) {
                        mx_f = cnt;
                        mx_f_arg = SA.sa[i-1];
                    }
                    i = j-1;
                }
            }
            if(mx_f == 0) {
                for(i = 0; i < SA.n; i++) {
                    if(SA.n - SA.sa[i] >= N) {
                        mx_f = 1;
                        mx_f_arg = SA.sa[i];
                        break;
                    }
                }
            }
            printf("%d ", mx_f);
            for(i = 0; i < N; i++)
                putchar(SA.str[mx_f_arg+i]);
            puts("");
        }
        while(getchar() != '\n');
    }
    return 0;
}