HDU 5907 Find Q-优快云博客

本文链接：https://blog.youkuaiyun.com/L_BestCoder/article/details/65445889

本文介绍了一种算法问题，即在一个长字符串中寻找所有仅由字母'q'组成的连续子串，并提供了一个高效的O(n)解决方案。通过遍历字符串并记录'q'出现的位置，可以计算出所有符合条件的子串数量。

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Find Q

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 1384 Accepted Submission(s): 608

Problem Description

Byteasar is addicted to the English letter 'q'. Now he comes across a string

S consisting of lowercase English letters.

He wants to find all the continous substrings of

S , which only contain the letter 'q'. But this string is really really long, so could you please write a program to help him?

Input

The first line of the input contains an integer

T(1≤T≤10) , denoting the number of test cases.

In each test case, there is a string

S , it is guaranteed that

S only contains lowercase letters and the length of

S is no more than

100000 .

Output

For each test case, print a line with an integer, denoting the number of continous substrings of

S , which only contain the letter 'q'.

Sample Input

  
  
   
   2
qoder
quailtyqqq

Sample Output

1
7

题意：给出一个字符串，求该字符串有多少只含q字母的连续子串。

思路：O(n)复杂度遍历，当遇到一个第一个q时，st标记起点，一直往后走，只到发现不为q的字母，然后用(len + 1) * len / 2 公式求得该子串包含有多少个子串。注意处理边界情况。

代码如下：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
char s[100005];
int main(){
	int cases, i, st, n, len, flag;
	long long sum, x;//注意数据范围 
	cin>>cases;
	while(cases--){
		scanf("%s", s);
 		len = strlen(s);
 		flag = false;//若已经发现了起点，则flag=true 
 		sum = 0;
 		for(i = 0; i < len; i++){
			if(s[i] == 'q' && flag == false){//发现起点 
				st = i;
				flag = true;
			}
			if(s[i] != 'q' && flag){//在发现起点的情况下开始找到了不是q的字符 
				x = i - st;
				sum += (1 + x) * x / 2;
				flag = false;
			}
			if(s[i] == 'q' && i == len - 1 && flag) {//若字符串最后一个字符为q 
				x = i - st + 1;
				sum += (1 + x) * x / 2;
			}
	 	}
	 	cout<<sum<<endl;
	}
	return 0;
}