Educational Codeforces Round 5 D. Longest k-Good Segment（双指针）

time limit per test
1 second
memory limit per test256 megabytes
The array a with n integers is given. Let’s call the sequence of one or more consecutive elements in a segment. Also let’s call the segment k-good if it contains no more than k different values.
Find any longest k-good segment.
As the input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printfinstead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input
The first line contains two integers n, k (1 ≤ k ≤ n ≤ 5·105) — the number of elements in a and the parameter k.
The second line contains n integers ai (0 ≤ ai ≤ 106) — the elements of the array a.
Output
Print two integers l, r (1 ≤ l ≤ r ≤ n) — the index of the left and the index of the right ends of some k-good longest segment. If there are several longest segments you can print any of them. The elements in a are numbered from 1 to n from left to right.
Examples
input
5 5
1 2 3 4 5
output
1 5
input
9 3
6 5 1 2 3 2 1 4 5
output
3 7
input
3 1
1 2 3
output
1 1

意思是说给你一个数组，然后再给你一个K，让你求出一个区间，这个区间中不同的数字最多只有K种，并且是最长的一个。

用两个变量来表示左右区间，比如用l和r来代表左右区间，然后不断更新r，只有当满足条件了，才开始更新l，每次更新r的时候都要判断这个数字是否出现过，这个好办，只要用个标记数组标记就行，毕竟数字并不是很大，之所以只有在满足条件的时候才更新l，是因为如果只是改变l，那只会改变一个数字在区间中出现的次数，新的l到r之间的数字是已经统计好了的，没必要重新判断，这样就可以累加到最后。

#include
#include
#include
#include
#include
#include
#include
#include
#define maxn 500010
#define maxm 6000100
using namespace std;
int arr[maxn];
int f[maxm];
int n,k;
int main()
{
    scanf("%d %d",&n,&k);
    for(int i=0;iscanf("%d",&arr[i]);
    int l=0;
    int r=0;
    int kind=0;
    int ans=0;
    int nut=0;
    int ansl=0;
    int ansr=0;
    while(rwhile(kindif(rif(!f[arr[r]]) kind++; 
                f[arr[r]]++; 
                r++;
            }
            else  break;    
        }
        while(kind==k)
        {
            if(relse  break;
        }
        ans=r-l;
        if(ans>nut)
        {
            ansl=l+1;
            ansr=r;
            nut=ans;
        }
        f[arr[l]]--;
        if(f[arr[l]]==0)  kind--;
        l++;
    }
    printf("%d %d\n",ansl,ansr);

    return 0;
}