HDU - 5672 String

There is a string S only contain lower case English character.(10≤length(S)≤1,000,000)
How many substrings there are that contain at least k(1≤k≤26) distinct characters?
Input
There are multiple test cases. The first line of input contains an integer T(1≤T≤10) indicating the number of test cases. For each test case:

The first line contains string S.
The second line contains a integer k(1≤k≤26).
Output
For each test case, output the number of substrings that contain at least k dictinct characters.
Sample Input
2
abcabcabca
4
abcabcabcabc
3
Sample Output
0
55
題目大意： 給最長1e6的字符串和k，讓你求有多少個子串，每個子串至少包含k個不同的字符。
思路: 尺取：既然讓找的每個子串至少包含k個不同的字符，就設兩個變量 l和r，從頭開始，固定 l，r後移，直到有出現k個不同的子串停止，則 r及右端點後面的不影響該子串，所以從 l開始的滿足要求的子串有 n-r+1個，如此遍歷每個左端點；
小優化：將沒出現過的字符標記一下，出現過則不統計，這樣r就可以不用一直從1開始。
ac代碼

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1e6+7;
const ll INF=1e9+7;
ll n;
char c[maxn];
ll vis[maxn];
int main()
{
    ll t;
    scanf("%lld",&t);
    while(t--)
    {
        memset(vis,0,sizeof(vis));
        scanf("%s",c);
        scanf("%lld",&n);
        ll l=0,r=0;
        ll len=strlen(c);
        ll k=0,ans=0;
        while(l<len)
        {
            //r=l;
            while(r<len&&k<n)
            {
                if(vis[c[r]-'a']==0)
                    k++;
                vis[c[r]-'a']++;
                r++;
            }
            if(k<n)             //如果剩下的不同字符少於最少要滿足的個數，退出
                break;
            ans+=len-r+1;
            vis[c[l]-'a']--;
            if(vis[c[l]-'a']==0)        
                k--;
            l++;
        }
        printf("%lld\n",ans);
    }
    return 0;
}