【HDU - 5056】Boring count (尺取法)
題干:
You are given a string S consisting of lowercase letters, and your task is counting the number of substring that the number of each lowercase letter in the substring is no more than K.
Input
In the first line there is an integer T , indicates the number of test cases.?
For each case, the first line contains a string which only consist of lowercase letters. The second line contains an integer K.?
[Technical Specification]?
1<=T<= 100?
1 <= the length of S <= 100000?
1 <= K <= 100000
Output
For each case, output a line contains the answer.
Sample Input
3 abc 1 abcabc 1 abcabc 2Sample Output
6 15 21題目大意:
找出字符串中所有字母出現(xiàn)次數(shù)不超過(guò)k次的子串?dāng)?shù)量
解題報(bào)告:
? ? 尺取就可以了。
AC代碼:
#include<bits/stdc++.h> #define ll long long using namespace std;char a[100000 + 5]; int bk[40]; int k; ll ans; int main() { // freopen("in4.txt","r",stdin);int t,l,r,len;cin>>t;while(t--) {memset(bk,0,sizeof(bk) );ans = 0; scanf("%s",a);scanf("%d",&k);len = strlen(a);l=r=0;while(r<len) {bk[a[r]-'a']++;while(bk[ a[r] - 'a' ] > k) {//注意這是while不是ifbk[ a[l] - 'a']--;l++;} // printf("%lld %d %d \n",ans,l,r);ans+=r-l+1;r++; }printf("%lld\n",ans);} /** //以左端點(diǎn) while(r<len) {if(l == r) {r++;}}//后續(xù)處理 while(l<=r) {} */ return 0 ; }?
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