pat 1093. Count PAT's (25)

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:

2

解：dp法，P[i]表示 到i位置前P的个数；PA[i]表示 到i位置前PA的个数；PAT[i]表示 到i位置前PAT的个数，状态转移条件代码里。

代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
    string s;
    cin>>s;
    int len=s.length();
    int *P=new int[len+10];
    int *PA=new int[len+10];
    int *PAT=new int[len+10];
    memset(P,0,sizeof(P));
    memset(PA,0,sizeof(PA));
    memset(PAT,0,sizeof(PAT));
    for(int i=0;i<len;i++)
    {
        if(s[i]=='P')
        {
            if(i==0)
                P[i]=1;
            else
                P[i]=P[i-1]+1;
        }
        else
        {
           if(i==0) P[i]=0;
           else P[i]=P[i-1];
        }
    }
    for(int i=0;i<len;i++)
    {
        if(s[i]=='A'&&P[i]!=0)
        {
            if(i==0)
                PA[i]=1;
            else
            {
                PA[i]=PA[i-1]+P[i];
                PA[i]%=1000000007;       
            }
        }
        else
        {
           if(i==0) PA[i]=0;
           else PA[i]=PA[i-1];
        }
    }
    for(int i=0;i<len;i++)
    {
        if(s[i]=='T'&&PA[i]!=0)
        {
            if(i==0)
                PAT[i]=1;
            else
             {
                PAT[i]=PAT[i-1]+PA[i];
                PAT[i]%=1000000007;   
             }  
        }
        else
        {
           if(i==0) PAT[i]=0;
           else PAT[i]=PAT[i-1];
        }
    }
    cout<<PAT[len-1]<<endl;
}

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