Broken Keyboard（hash）

本是古典何须时尚 2022-08-02 06:22 22阅读 0赞

# Broken Keyboard #

**Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 691    Accepted Submission(s): 274**

Problem Description

Bruce Force's keyboard is broken, only a few keys are still working. Bruce has figured out he can still type texts by switching the keyboard layout whenever he needs to type a letter which is currently not mapped to any of the m working keys of the keyboard.   
  
You are given a text that Bruce wishes to type, and he asks you if you can tell him the maximum number of consecutive characters in the text which can be typed without having to switch the keyboard layout. For simplicity, we assume that each key of the keyboard will be mapped to exactly one character, and it is not possible to type other characters by combination of different keys. This means that Bruce wants to know the length of the largest substring of the text which consists of at most m different characters.

Input

The input contains several test cases, each test case consisting of two lines. The first line of each test case contains the number m (1 ≤ m ≤ 128), which specifies how many keys on the keyboard are still working. The second line of each test case contains the text which Bruce wants to type. You may assume that the length of this text does not exceed 1 million characters. Note that the input may contain space characters, which should be handled like any other character.   
  
The last test case is followed by a line containing one zero.

Output

For each test case, print one line with the length of the largest substring of the text which consists of at most m different characters.

Sample Input

5
    This can't be solved by brute force.
    1
    Mississippi
    0

Sample Output

7
    2
    
       
        Hint The largest substring for the first test case is "_by_bru", where _ stands for a space character.

Source

[2008水题公开赛（比速度，OJ压力测试）][2008_OJ]

**题意很简单就是让你用·n个不同的字符组成的最大字符串的长度！**

#include <iostream>
    #include<cstring>
    using namespace std;
    char key[1100000];
    long hash1[256];
    int main()
    {
        long n,num,p;
        while (scanf("%ld",&n)!=EOF&&n)
        {
            getchar();
            gets(key);
            memset(hash1,0,sizeof(hash1));
            long i,start;
            long diff=0;
            long res=0;
            for (start=i=0;key[i]!='\0';++i)
            {
                ++hash1[key[i]];//映射值+
                if (hash1[key[i]]==1)//如果是第一次出现
                {
                    ++diff;//不同字母数+1
                    if (diff>n)//如果超过n个不同字母
                    {
                       /* c=key[start];
                        while(key[start]==c)
                        {
                            hash[c]--;
                            start++;
                        }
                        比如：aabacdefg n=3
                        最大是5当程序运行到d的时候如何消掉a呢？我用上面那个方法发现消不掉a因为中间隔了一个b
                        那样的话diff不会减一！
                        别人的想法是把b消掉，就是这样做的！
                        上面这个有问题的，我当时的意思是*/
                        /*while(hash[key[start]]>0)
                        {
                            hash[key[start]]--;
                            start++;
                        }
                        上面这个也不对，为什么呢？
                        比如abcde n=4我只是想消掉a但是你会发现都消完了，所以上面那个不对！*/
                        do
                        {
                            hash1[key[start]]--;
                            start++;
                        }while(hash1[key[start-1]]>0);//这一点需要判断一下前面那个是否为0！
                       diff--;
                    }
                }
                long len=i-start+1;//得到此时的串长度
                res=res>len?res:len;//保存最大值
            }
            printf("%ld\n",res);
        }
        return 0;
    }

[2008_OJ]: http://acm.hdu.edu.cn/search.php?field=problem&key=2008%CB%AE%CC%E2%B9%AB%BF%AA%C8%FC%A3%A8%B1%C8%CB%D9%B6%C8%A3%ACOJ%D1%B9%C1%A6%B2%E2%CA%D4%A3%A9&source=1&searchmode=source