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.
Each input file contains one test case. For each case, there is only one line giving a string of no more than 1 characters containing only P, A, or T.
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.
APPAPT
2计数方法，先计数T的个数，再遍历P增加，T减少，遇到一个A进行累加
#include <iostream> using namespace std; int main() { string s; getline(cin,s); long long P=0,A=0,T=0,sum=0; for(int i=0;i<s.length();i++) if(s[i]==‘T‘) T++; for(int i=0;i<s.length();i++){ if(s[i]==‘P‘) P++; if(s[i]==‘T‘) T--; if(s[i]==‘A‘) sum+=((P%1000000007*T%1000000007)%1000000007); } cout<<sum%1000000007; system("pause"); return 0; }
PAT Advanced 1093 Count PAT's (25分)
原文：https://www.cnblogs.com/littlepage/p/12216152.html