interviewstreet-string similarity – 類別-string process

題目來源：https://www.interviewstreet.com/challenges/dashboard/#problem/4edb8abd7cacd

解題報告：

簡單的字串處理，對S的每個尾碼，計算它與S的相似性，相似性計算是用遞迴做的。

#include <iostream>#include <string.h>using namespace std;int getSim(char* str1, char* str2){    if ( *str1 == '\0' || *str2 == '\0')        return 0;    if (*str1 != *str2)        return 0;    if ( *str1 == *str2)        return 1 + getSim(str1+1, str2+1);}int main(){    char * s = new char[100000];    int N;    cin >> N;    for (int i = 0; i < N; i++)    {        int similarity = 0;        cin >> s;        similarity += strlen(s);        for (int i = 1; i < strlen(s); i++)        {            similarity += getSim(s, s + i);        }        cout << similarity << endl;    }}

附錄：

String Similarity (25 Points)

For two strings A and B, we define the similarity of the strings to be the length of the longest prefix common to both strings. For example, the similarity of strings "abc" and "abd" is 2, while the similarity of strings "aaa" and "aaab" is 3.

Calculate the sum of similarities of a string S with each of it's suffixes.

Input:
The first line contains the number of test cases T. Each of the next T lines contains a string each.

Output:
Output T lines containing the answer for the corresponding test case.

Constraints:
1 <= T <= 10
The length of each string is at most 100000 and contains only lower case characters.

Sample Input:
2
ababaa
aa

Sample Output:
11
3

Explanation:
For the first case, the suffixes of the string are "ababaa", "babaa", "abaa", "baa", "aa" and "a". The similarities of each of these strings with the string "ababaa" are 6,0,3,0,1,1 respectively. Thus the answer is 6 + 0 + 3 + 0 + 1 + 1 = 11.

For the second case, the answer is 2 + 1 = 3.