例题8-15(uva-12174)

#include <iostream>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <numeric>
#include <chrono>
#include <ctime>
#include <cmath>
#include <cctype>
#include <string>
#include <cstdio>
#include <iomanip>


#include <thread>
#include <mutex>
#include <condition_variable>
#include <functional>
#include <iterator>
using namespace std;
const int maxn = 1e5 + 7;
int s, n,t,input[maxn], tcnt[maxn] = {
    
    0};
char res[maxn];

int main()
{
    
    
	cin >> t;
	while (t--) {
    
    
		cin >> s >> n;
		memset(tcnt, 0, sizeof(tcnt));
		memset(res, 1, sizeof(res));
		int ans = 0, oneCnt = 0;
		for (int i = 0; i < n; i++) {
    
    
			cin >> input[i];
		}
		
		for (int i = 0; i < n + s; i++) {
    
    
			//新出现的元素
			if(i < n) oneCnt += ++tcnt[input[i]] == 1;
			//消失的元素
			if(i >= s)	oneCnt -= --tcnt[input[i - s]] == 0;
			//当前窗口长度 与出现元素个数相同 则每个元素都是一个 不是则有重复元素
			//不能以i作为窗口结尾
			if (min(n - 1, i) - max(i - s + 1, 0) + 1 != oneCnt) res[i%s] = 0;
		}

		for (int i = 0; i < s; i++) {
    
    
			if (res[i]) ++ans;
		}
		cout << ans << endl;
	}
	return 0;
}