HDU 5056 Boring count(不超过k个字符的子串个数)
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Boring count
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 828 Accepted Submission(s): 342
Problem Description
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
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
3abc1abcabc1abcabc2
Sample Output
61521
Source
BestCoder Round #11 (Div. 2)
解题思路:枚举字符串下标i,每次计算以i为结尾的符合条件的最长串。那么以i为结尾的符合条件子串个数就是最长串的长度。求和即可。
代码如下:
#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <string>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <numeric>#include <iomanip>#include <bitset>#include <sstream>#include <fstream>#include <limits.h>#define debug "output for debug\n"#define pi (acos(-1.0))#define eps (1e-6)#define inf (1<<28)#define sqr(x) (x) * (x)#define mod 1000000007using namespace std;typedef long long ll;typedef unsigned long long ULL;#define maxn 100010int n,m,t,k;char str[maxn];int num[maxn];int main(){ scanf("%d",&t); while(t--) { memset(num,0,sizeof(num)); scanf("%s",str); scanf("%d",&k); int k1=strlen(str); int pos=0; ll ans=0; for(int i=0; i<k1; i++) { num[str[i]-'a']++; if(num[str[i]-'a']>k) { while(str[pos]!=str[i]) { num[str[pos]-'a']--; pos++; } num[str[pos]-'a']--; pos++; } ans+=(i-pos+1); } printf("%I64d\n",ans); } return 0;}
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