HDU5907-Find Q

最新推荐文章于 2020-08-17 15:18:11 发布

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本文介绍了一道编程题目，任务是编写一个程序来计算给定字符串中只包含字母'q'的所有连续子串的数量。输入包括多组测试案例，每组案例是一个长度不超过100000的字符串。

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Find Q

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 319 Accepted Submission(s): 181

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

Source

BestCoder Round #88

Recommend

wange2014

题意：给出一串字符串S，问S中的仅包含字母q的连续子串有几个

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

const int INF=0x3f3f3f3f;

int main()
{
    int t;
    char ch[100090];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",ch);
        int i=0;
        long long int sum=0;
        while(ch[i])
        {
            if(ch[i]!='q') i++;
            else
            {
                long long int k=1;
                i++;
                while(ch[i]=='q')
                {
                    k++;
                    i++;
                }
                sum+=(k+1)*k/2;
            }
        }
        printf("%lld\n",sum);
    }
    return 0;
}