1093. Count PAT's (25)想法题吧，算是排列组合吧

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

提交代码

#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=200000;
const int inf=2100000000;
long long  mod=1000000007;
char s[maxn];
int a[maxn];
int b[maxn];
int main()
{
    int n,m,i,k,t,j;
    scanf("%s",s);
    n=strlen(s);
    long long  ans=0;
    int num=0;
    int cnt=0,cnt2=0;
    //统计A前面的P的个数
    for(i=0;i<n;i++)
    {
        if(s[i]=='P')num++;
        else if(s[i]=='A')
        {
            a[cnt++]=num;
        }

    }
    cnt2=cnt-1;num=0;//统计A后面的T的个数
    for(i=n-1;i>=0;i--)
    {
        if(s[i]=='T')num++;
        else if(s[i]=='A')
        {
            b[cnt2--]=num;
        }
    }

    for(i=0;i<cnt;i++)//计算总和
    {
        ans+=a[i]*b[i]%mod;
        ans%=mod;
    }

    printf("%lld\n",ans%mod);
    return 0;
}

1093. Count PAT's (25)想法题吧，算是排列组合吧