https://www.acwing.com/problem/content/1585/
状态机的解法
#include <iostream>
#include <cstring>
using namespace std;
const int N = 100010, MOD = 1e9 + 7;
int n;
char s[N], p[] = " PAT";
int f[N][4];
int main()
{
cin >> s + 1;
n = strlen(s + 1);
f[0][0] = 1;
for (int i = 1; i <= n; i ++ )
for (int j = 0; j <= 3; j ++ )
{
f[i][j] = f[i - 1][j];
if (s[i] == p[j]) f[i][j] = (f[i][j] + f[i - 1][j - 1]) % MOD;
}
cout << f[n][3] << endl;
return 0;
}
前缀和+状态机的思想 的解法
#include <iostream>
using namespace std;
const int mod = 1e9 + 7;
long long p, a, b, res;
int main()
{
string str;
cin >> str;
for (auto c : str)
if (c == 'P') p ++;
else if (c == 'A') b += p;
else res = (res + b) % mod;
cout << res;
}
标签:PAT,int,res,状态机,计数,str,include
From: https://www.cnblogs.com/xjtfate/p/16614238.html