42: Maximum number of books published, 42:

42: Maximum number of books published, 42:
42: the largest number of books published

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Total time limit:

1000 ms

 

Memory limit:

65536kB

Description

Assume that the Library is new to m (10 ≤ m ≤ 999) books, which are prepared independently or jointly by n (1 ≤ n ≤ 26) authors. Assume that the number of m books is an integer (1 to 999) and the author's name is A letter ('A' to 'Z '), find out the authors who are most involved in the compilation and their book lists based on the book author list.

Input

In the first action, the number of books is m, and in the other m rows, each row is the information of a book. The first integer is the book number, A space is followed by a string consisting of uppercase letters without repeated characters. Each letter represents an author. Input data ensures that only one author has the most data published.

Output

The output has multiple rows:
The first act is the author letter with the largest number of books published;
The number of books published by the second act author;
The number of the books that the authors of other behaviors participate in (output in the input order ).

Sample Input

11307 F895 H410 GPKCV567 SPIM822 YSHDLPM834 BXPRD872 LJU791 BPJWIA580 AGMVY619 NAFL233 PDJWXK

Sample output

P6410567822834791233

Source

Introduction to computing of Peking University 06 psychological and Information Management Final Examination

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 using namespace std; 5 int a [10001]; 6 int now; 7 int sl [10001]; 8 struct node 9 {10 int a; // number 11 char B [1001]; // author 12 int lb; 13} gs [1001]; 14 int maxn =-1; 15 char ans; 16 int main () 17 {18 int n; 19 cin> n; 20 for (int I = 1; I <= n; I ++) 21 {22 cin> gs [I]. a; 23 // gets (gs [I]. b); 24 scanf ("% s", & gs [I]. b); 25 for (int j = 0; j <strlen (gs [I]. b); j ++) 26 {27 sl [gs [I]. B [j] ++; 28 if (sl [gs [I]. B [j]> maxn) 29 {30 ans = gs [I]. B [j]; 31 maxn = sl [gs [I]. B [j]; 32} 33} 34} 35 cout <ans <endl; 36 cout <maxn <endl; 37 for (int I = 1; I <= n; I ++) 38 {39 for (int j = 0; j <strlen (gs [I]. b); j ++) 40 {41 if (gs [I]. B [j] = ans) 42 {43 cout <gs [I]. a <endl; 44} 45} 46} 47 48 return 0; 49}