HDU 5907 Find Q

最新推荐文章于 2019-07-29 18:35:42 发布

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最新推荐文章于 2019-07-29 18:35:42 发布

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本文介绍了一种算法，用于在长字符串中查找所有连续的仅包含字母'q'的子串，并通过示例输入输出展示了程序的效果。

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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

很无语，我用自己的错误之路得到一个经验，一定要保证类型相同，否则就是交不上；

一开始，l与r的类型是int，因为它们最大不会超过100000，但是sum类型是long long类型的，最后

把l与r的类型改成long long的就AC了。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
int main()
{
    char s[110000];
    int t,i;
    __int64 l,r;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",s);
        int len=strlen(s);
        l=0;
        r=0;
        __int64 sum=0;
        for(i=0;i<len;i++)
        {
            if(s[i]=='q')
                r++;
            else
            {
                sum+=(r-l)*(1+r-l)/2;
                l=i;
                r=i;
            }
        }
        sum+=(r-l)*(1+r-l)/2;
        printf("%I64d\n",sum);
    }
    return 0;
 }