Daydream

描述

Welcome to 2009 HDU Girl’s Cup, bless you will happy in it.

Every girls are beautiful if you use you heart to feel. Every corner in the world will colourful and energetic by several girls standing. If you boy, a normal bay, I believe that you will try to watch when a beautiful girl passing you and you will nervous if a girl watching you just when you are watching her.

Now give you a surprise that may be never happy in the real time. Millions of PLMM stand in a line and facing you(^_^). They are dress colourful clothings. You should to find out a maximal subline PLMM that their clothing color are all different.

输入

The input contains multiple test cases.
Each case first give a integer n expressing many many girls stand in line.(n<=10000000)
Next line a string including n character, each character standing one girls’s clothing color.

输出

Output one integer the numbers of maximal subline PLMM that their clothing color are all different and the line's begin and end (based from 0). If their are more answer, print the begin smallest.

样例输入

3
abc
5
aaaba
8
hdugirls

样例输出

3 0 2
2 2 3
8 0 7

和吉首大学热身赛的题目，当时题意没看懂，现在知道了，就是求最长不重复子序列，输出他的长度、初始及结束位置

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
char str[10000005];
int main()
{
    int a[300],i;
    int n,mx,mi,x,s;
    while(~scanf("%d",&n))
    {
        scanf("%s",str);
        memset(a,0,sizeof(a));
        mx=mi=x=s=i=0;
        a[str[i]]=1;
        while(str[i]){
            i++;
            if(a[str[i]]==0)
                 a[str[i]]=1;
            else
            {
                x=i-1;
                if(x-s>mx-mi){
                    mx=x;
                    mi=s;
                }
                for(int j=s;j<=x;j++){
                    if(str[j]==str[i]){
                        s = j+1;
                        break;
                    }
                    else
                        a[str[j]]=0;
                }
            }
        if(i-1-s>mx-mi)
        {
           mx=i-1;
           mi=s;
        }
       }
      printf("%d %d %d\n",mx-mi+1,mi,mx);
    }
    return 0;
}