1093 Count PAT's （25 分）(逻辑题，水题)

最新推荐文章于 2020-02-11 17:10:42 发布

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最新推荐文章于 2020-02-11 17:10:42 发布

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分类专栏： c++ 甲级文章标签： PAT

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本文介绍了一种算法，用于计算给定字符串中特定子串“PAT”的出现次数。通过遍历字符串并利用计数变量，算法能高效地找到所有符合条件的子串组合。输入为包含P、A、T字符的字符串，输出为子串PAT的总数，结果模1000000007。

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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

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.

Sample Input:

APPAPT

Sample Output:

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <cctype>
#include <string.h>
#include <cstdio>
#include <unordered_set>
using namespace std;
int main() {
    string s;
    cin>>s;
    int len=s.length(),cntt=0,cntp=0,sum=0;
    for(int i=0;i<len;i++){
        if(s[i]=='T')
            cntt++;
    }
    for(int i=0;i<len;i++){
        if(s[i]=='P') cntp++;
        if(s[i]=='T') cntt--;
        if(s[i]=='A') sum=(sum+cntt*cntp)%1000000007;
    }
    cout<<sum;
    return 0;
}