【PTA-乙级】1040.有几个PAT

超时解法= =

#include<iostream>
using namespace std;
char arr[100010];

int main()
{
    long long num = 0;
    cin >> arr;
    char* p = arr;
    char* tmp1 = p;
    char* tmp2 = p;
    char* tmp3 = p;
    int flag = 1;
    while (*p != '\0')
    {
        tmp1 = p;
        if (*tmp1 == 'P')
        {
            int flag = 0;
            p++;
            //   AP PA APTTATTAAATATATTTAAT
            for (tmp2 = tmp1 + 1; *tmp2 != '\0'; tmp2++)
            {
                if (*tmp2 == 'A')
                {
                    for (tmp3 = tmp2 + 1; *tmp3 != '\0'; tmp3++)
                    {
                        if (*tmp3 == 'T')
                        {
                            flag = 1;
                            num++;
                        }
                    }
                }
            }
        }
        else
        {
            p++;
            continue;
        }
        if (!flag)
        {
            break;
        }
    }
    cout << num % 1000000007;
    return 0;
}

某柳姓大神的解法（实在NB）

#include<iostream>
using namespace std;
int main()
{
	string s;
	cin >> s;
	int len = s.length();
	long long countp = 0, countt = 0;
	long long num = 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')
			num += countp * countt;
	}
	cout << num % 1000000007;
	return 0;
}

第三种AC解法：

#include<iostream>
#include<cstring>
using namespace std;
char arr[100010];
int main()
{
	cin >> arr;
	int len = strlen(arr);
	long long t = 0, at = 0, pat = 0;
	for (int i = len - 1; i >= 0; i--)
	{
		if (arr[i] == 'T')
			t++;
		if (arr[i] == 'A')
			at += t;
		if (arr[i] == 'P')
			pat += at;
	}
	cout << pat % 1000000007;
	return 0;
}