1093 Count PAT's (25 分)

最新推荐文章于 2021-01-09 23:53:25 发布

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最新推荐文章于 2021-01-09 23:53:25 发布

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分类专栏： PAT

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本文介绍了一种用于计算字符串中特定子串'PAT'出现次数的高效算法。通过使用递推方式，预先计算出每个字符左侧的'P'数量和右侧的'T'数量，从而在O(n)时间内得出结果。文章提供了完整的C++代码实现。

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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
5
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

递推算法：每一位要计算的值都可以通过两边的计算结果获得
题中只需要计算每一个‘A’两边有多少个‘P’和‘T’相乘就是总数。使用数组Lp记录每一个下标为i的位置左边有多少个P。右边不用记录每个位置右边T的个数，可以使用一个变量储存当前访问的位置左边有几个T，当访问到A时直接进行计算结果ans。
注意越界，写法：ans=(ans+Rt*Lp[i])%mode

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn=100010;
const int mode=1000000007;
char c[maxn];
int Lp[maxn];
int main(){
	scanf("%s",&c);
	int len = strlen(c);
	for(int i=0;i<len;i++){
		if(i>0)
			Lp[i]=Lp[i-1];
		if(c[i]=='P')
			Lp[i]++;	
	}
	int ans=0,Rt=0;
	for(int i=len-1;i>0;i--){
		if(c[i]=='T')
			Rt++;
		else if(c[i]=='A'){
			ans=(ans+Rt*Lp[i])%mode;
		}	
	} 
	cout << ans <<endl;
	return 0;
}