pat1093. Count PAT's (25)

最新推荐文章于 2020-06-20 10:58:54 发布

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最新推荐文章于 2020-06-20 10:58:54 发布

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原文链接：http://www.cnblogs.com/Deribs4/p/4762128.html

本文介绍了一种算法，用于计算给定字符串中特定子串PAT出现的次数，并提供了一个C++实现示例。该算法通过遍历输入字符串并利用模运算来避免溢出，确保了即使对于非常大的输入也能正确输出结果。

1093. Count PAT's (25)

时间限制

120 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CAO, Peng

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 10⁵characters 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.

Sample Input:

APPAPT

Sample Output:

提交代码

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<queue>
 6 #include<vector>
 7 #include<cmath>
 8 using namespace std;
 9 #define mod 1000000007
10 int main(){
11     string s;
12     cin>>s;
13     int countP=0,countPA=0,count=0;
14     int i;
15     for(i=0;i<s.length();i++){
16         if(s[i]=='P'){
17             countP++;
18             countP=countP%mod;
19         }else{
20             if(s[i]=='A'){
21                 countPA+=countP;
22                 countPA=countPA%mod;
23             }
24             else{
25                 count+=countPA;
26                 count=count%mod;
27             }
28         }
29     }
30     cout<<count<<endl;
31     return 0;
32 }