PAT (Advanced Level) Practise 1093 Count PAT's (25)

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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'出现的次数。通过预处理字符计数并利用这些计数进行计算，避免了直接匹配子串所带来的复杂度。

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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 10⁵ 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:

题意：给你一个字符串，问一共有几个“PAT”

解题思路：算出每个A左边有几个P，右边有几个T，乘一下就好

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <cmath>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

#define mod 1000000007

char ch[100005];
int cnt1[100005],cnt2[100005];

int main()
{
    while(~scanf("%s",ch+1))
    {
        int len=strlen(ch+1);
        cnt1[0]=0;
        for(int i=1;i<=len;i++)
        {
            cnt1[i]=cnt1[i-1];
            if(ch[i]=='P') cnt1[i]++;
        }
        cnt2[len+1]=0;
        for(int i=len;i>=1;i--)
        {
            cnt2[i]=cnt2[i+1];
            if(ch[i]=='T') cnt2[i]++;
        }
        LL ans=0;
        for(int i=1;i<=len;i++)
            if(ch[i]=='A') ans=(ans+1LL*(cnt1[i]*cnt2[i])%mod)%mod;
        printf("%lld\n",ans);
    }
    return 0;
}