1093 Count PAT’s
part 3, 3.5
自己解法
- 本来说是啥活用递归，但是我不懂为啥要用递归hhh，我的思路和柳神一模一样，他的还要更简洁
#include <iostream>
using namespace std;
#include <vector>
#include <algorithm>int main()
{string s;cin >> s;long long count = 0, cP = 0, cA = 0, cT = 0;for (int i = 0; i < s.size(); i++)if (s[i] == 'P')cP++;else if (s[i] == 'A')cA++;elsecT++;long long iP = 0, iT = 0;for (int i = 0; i < s.size(); i++)if (s[i] == 'P')iP++;else if (s[i] == 'T')iT++;elsecount = (count + iP * (cT - iT)) % 1000000007;cout << count << endl;system("pause");return 0;
}
大神解法
- 柳神
#include <iostream>
#include <string>
using namespace std;
int main() {string s;cin >> s;int len = s.length(), result = 0, countp = 0, countt = 0;for (int i = 0; i < len; i++) {if (s[i] == 'T')countt++;}for (int i = 0; i < len; i++) {if (s[i] == 'P') countp++;if (s[i] == 'T') countt--;if (s[i] == 'A') result = (result + (countp * countt) % 1000000007) % 1000000007;}cout << result;return 0;
}