HDU - 5056（尺取）

最新推荐文章于 2021-01-06 02:16:56 发布

XiWenjuan086

最新推荐文章于 2021-01-06 02:16:56 发布

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分类专栏：算法杂烩

本文链接：https://blog.youkuaiyun.com/weixin_43871207/article/details/88410034

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本文详细解析了Boringcount问题，这是一个关于字符串中特定子串计数的算法挑战。通过滑动窗口技术，我们实现了高效的解决方案，适用于大量数据集。文章提供了完整的C++代码实现，展示了如何在限定条件下计算字符串中所有子串中每个小写字母出现次数不超过K的子串数量。

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Boring count

Problem Description
You are given a string S consisting of lowercase letters, and your task is counting the number of substring that the number of each lowercase letter in the substring is no more than K.
Input In the first line there is an integer T , indicates the number of test cases.
For each case, the first line contains a string which only consist of lowercase letters. The second line contains an integer K.

[Technical Specification]
1<=T<= 100
1 <= the length of S <= 100000
1 <= K <= 100000
Output For each case, output a line contains the answer.
Sample Input3
abc
1
abcabc
1
abcabc
2
Sample Output6
15
21

尺取法。在网上搜的题解

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
const int maxn=1e5+5;
int k,len,cnt[200];
char s[maxn];
bool check()
{
    for(char i='a';i<='z';i++)
        if(cnt[i]>k)
            return false;
    return true;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(cnt,0,sizeof(cnt));
        cin>>s;
        scanf("%d",&k);
        len=strlen(s);
        long long ans=0;
        for(int i=0,j=0;i<len;i++)
        {
            cnt[s[i]]++;
            while(check()==false)
            {
                cnt[s[j]]--;
                j++;
            }
            ans+=i-j+1;
        }
        cout<<ans<<'\n';
    }
    return 0;
}