1093 Count PAT's

1093 Count PAT‘s

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 10?5?? 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.

思路：对于每个A，记录它的前面有几个P，后面有几个T；然后对于每个A，PAT的组合有（前P）*（后T）种

 1 #include <iostream>
 2 #include <string>
 3 #include <vector>
 4 #include <queue>
 5 using namespace std;
 6 const int maxn = 100005;
 7 
 8 int main(){
 9     string s; cin >> s;
10     int pre[maxn], back[maxn];
11     fill(pre,pre+maxn,0); fill(back, back+maxn, 0);
12     int preP=0;
13     for(int i=0;i<s.length();i++){
14         pre[i] = preP;
15         if(s[i]==‘P‘) preP++;
16     }
17     int backT=0;
18     for(int i=s.length()-1;i>=0;i--){
19         back[i] = backT;
20         if(s[i]==‘T‘) backT++;
21     }
22     long long cnt;
23     for(int i=0;i<s.length();i++){
24         if(s[i]==‘A‘){
25             cnt += pre[i]*back[i];
26         }
27         if(cnt>1000000007) cnt%=1000000007;
28     }
29     printf("%lld\n",cnt);
30 }

1093 Count PAT's

原文：https://www.cnblogs.com/lokwongho/p/9940857.html