【PAT甲级真题】- Count PAT‘s (25)

题目来源

Count PAT's (25)

注意点

• 结果要取模

题目描述

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.

输入描述:

Each input file contains one test case. For each case, there is only one line giving a string of no more than 10 5 10^5 105

characters containing only P, A, or T.

输出描述:

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.

输入例子:

APPAPT

输出例子:

2

思路简介

一道经典的 DP 题

推理一遍就是：

1. 如果当前 "PA" 的数目是 s1，"PAT" 的数目是 s2，那么输入一个 "T" ，它可以和前面所有的 "PA" 组成新的 "PAT" ，总的 "PAT" 就是新的加上旧的，即 s1+s2
2. 同理如果当前 "P" 的数目是 s0，"PA" 的数目是 s1，那么输入一个 "A" ，它可以和前面所有的 "P" 组成新的 "PA" ，新的加上旧的即 "PA" 的总数

转移方程就是

if cin==T:
	PAT=PAT+PA
else if cin== A:
	PA=PA+P
else:
	P++

遇到的问题

无，一遍过

代码

/**
 * https://www.nowcoder.com/pat/5/problem/4039
 * dp
 */
#include<bits/stdc++.h>
using namespace std;


void solve(){
    string s;
    cin>>s;
    int len=s.size();
    int p=0,pa=0,pat=0;
    for(int i=0;i<len;++i){
        if(s[i]=='P')p++;
        if(s[i]=='A')pa=pa+p;
        if(s[i]=='T')pat=(pa+pat)%1000000007;
    }
    cout<<pat;
}

int main(){
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    //fstream in("in.txt",ios::in);cin.rdbuf(in.rdbuf());
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}