HDOJ 5672 String

String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 965    Accepted Submission(s): 313

Problem 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?

 

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
   
   
   
   

 

中文题面： BC Round #81 (div.2)1003  

题解：额，组合数学和模拟被学弟吊打(；′⌒`)  /(ㄒoㄒ)/~~  想多了，还以为这一题要DP计数，感觉不会搞。 学弟直接模拟计数。我这组合数学太渣渣。。。 

 

按顺序遍历字符串，记录出现的不同字母的个数cnt，和每种字母出现的数量。 当cnt==k时，以当前字符串为子串的子串共有len-i个。然后将最左边的减去，依次记录下去。

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s[1000010];
int num[30];
int main()
{
	int k,t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s%d",s,&k);
		int len=strlen(s);
		int cnt=0,left=0;
		long long ans=0;
		memset(num,0,sizeof(num));
		for(int i=0;i<len;++i)
		{
			int temp=s[i]-'a';
			if(num[temp])
				num[temp]++;
			else
			{
				num[temp]++;
				cnt++;
			}
			if(cnt==k)
			{
				ans+=(len-i);
				while(1)
				{
					temp=s[left]-'a';
					num[temp]--;
					left++;
					if(!num[temp])
					{
						cnt--;
						break;
					}
					else
						ans+=(len-i);
				}
			}
		}
		printf("%I64d\n",ans);
	}
	return 0;
}