1093. Count PAT's (25)

本文介绍了一个算法问题,即计算字符串中特定子串“PAT”的出现次数,并提供了一段C++代码实现。输入为一个不超过10^5长度的字符串,仅包含字符P、A、T,输出为该字符串中“PAT”作为子串出现的次数,结果取模1000000007。

result moded by 1000000007 就是 cout << pat % 1000000007;


The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed by the 3rd, the 4th, and the 6th characters.

Now given any string, you are supposed to tell the number of PAT's contained in the string.

Input Specification:

Each input file contains one test case. For each case, there is only one line giving a string of no more than 105 characters containing only P, A, or T.

Output Specification:

For each test case, print in one line the number of PAT's contained in the string. Since the result may be a huge number, you only have to output the result moded by 1000000007.

Sample Input:
APPAPT
Sample Output:
2

 


   
  1. #include <iostream>
  2. #include <string>
  3. using namespace std;
  4. int main(void) {
  5. string s;
  6. cin >> s;
  7. long long int p=0, pa=0, pat=0;
  8. for (int i = 0; i < s.length(); i++) {
  9. if (s[i] == 'P')
  10. p++;
  11. else if (s[i] == 'A')
  12. pa += p;
  13. else
  14. pat += pa;
  15. }
  16. cout << pat % 1000000007;
  17. /*while (true)
  18. {
  19. }*/
  20. return 0;
  21. }





转载于:https://www.cnblogs.com/zzandliz/p/5023282.html

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