Description
There is a string S.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?
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).
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
对于给定长度的字符串要求出它的子串数目,对子串的要求是每个自子串中至少含有k个不同的字符。
这道题的数据量同样很大,我开始连暴力的想法都没有T_T。
听完学长讲题解,双指针(网上大家也都叫尺取法),好巧妙啊,觉得应该好难敲就好久都没实现,等好长时间以后写了一次,不知怎么一直RE,请教学长后,学到了这饿个很精炼的代码。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6+5;
char a[maxn];
int vis[100],T,n,len;
int f(){
int sum = 0;
for(int i=0;i<26;i++){
if(vis[i])
sum++;
}
return sum>=n;//已经找到了满足条件的一种子串了。
}
int main(){
scanf("%d",&T);
while(T--){
scanf("%s",a);
scanf("%d",&n);
memset(vis,0,sizeof(vis));
int l =0 ,r = 0;len=strlen(a);ll ans = 0;
while(r<len){
vis[a[r]-'a']++;
while(f()){
vis[a[l++]-'a']--;
}
ans+=l;// [l-1,r] 为最短答案,那么[0...l-1,e] 都是答案。所以,ans += l;
r++;
}
printf("%lld
",ans);
}
return 0;
}