题目描述
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.
输入
Each input file contains one test case. For each case, there is only one line giving a string of no more than $10^5 characters containing only P, A, or T.
输出
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.
思路
一道挺有意思的智力题。首先假设题目只让找到能组成P的个数，那么就是所有的P。接下来如果看找到PA的个数，可以发现我们从头遍历，对每一个P计数，当遇到A时，前面所有的P都可以形成PA因此都被记入。因此就是每到一个A我们记录它前面的P的个数，最后加起来即可，那么对于T呢，我们记入每一个PA的个数，遇到T加起来即可。
代码
#include<iostream>
#include<cstdio>
#include<stdlib.h>
using namespace std;
int main()
{
long long all = 0;
long long p = 0;
long long a = 0;
string str;
cin >> str;
for (int i = 0; i < str.size(); i++)
{
if (str[i] == 'P')
{
p++;
}
if (str[i] == 'A')
{
a = (a + p) % 1000000007;
}
if (str[i] == 'T')
{
all = (all + a) % 1000000007;
}
}
printf("%lld\n", all);
}