| # | Problem | Pass Rate (passed user / total user) |
|---|---|---|
| 11170 | The number of occurrences |
|
| 13309 | How much is the string worth 2 |
|
Description
Given a string A and n strings B1, B2, B3, …, Bn, count the number of occurrences of string A in each of B1, B2, B3, … , and print the maximum number of occurrences. All of the strings contain only digits.
For example, if A is “50” , n = 3, and the n strings are
“5005”
“055050”
“55000”
then your answer should be 2 because A appears in “5005” one time, in “055050” two times, and in “55000” one time. So the maximum number of occurrences is 2.
Note that if A is “99” and B1 is “9999”, the number of occurrences of A in B1 is counted as 3.
You may assume string B1 is always longer than string A.
Input
The first line of the input is the string A (0<length of A<=4). The second line is n (1<n<10) .
For the next n lines, each line contains a string Bi (length of A < length of Bi <9) and a ‘\n’ at the end of the line.
Output
The maximum number of occurrences of A appears in B1, B2, …, Bn. Note that you DO NOT need to print ‘\n’ at the end of the output.
Sample Input Download
Sample Output Download
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Description
In this problem, the first line of input contains a string C, we define C1 worth 1 coin, C2 worth 2 coins, C3 worth 3 coins ...., C26 worth 26 coins, other charactor worth 0 coin.
Now we have some lines of string S (may contain spaces), you need to write a program to calculate how much each line is worth until S = 0 or end.
0_ or _0 ( _ = space) worth 0, not the end of input
Input
The first line contains a string C, Ci worth i coins
The next lines, each line contains a string S (may contain spaces)
Testcases:
(4/7) C = abcdefghijklmnopqrstuvwxyz, 0 <= |S| <= 100, ∀S ≠ end
(1/7) C = abcdefghijklmnopqrstuvwxyz, 0 <= |S| <= 100
(1/7) C = the permutaion of a ~ z, 0 <= |S| <= 100, ∀S ≠ end
(1/7) C = the permutaion of a ~ z, 0 <= |S| <= 100
Output
Output the value of each line of string until S = 0 or end.