題意

求出一個字串中長度為$m$的子串有多少個不同的。

思路

$map$可拿50分。
字串雜湊，每次取出一段來判斷是否重複。單雜湊會被卡，這裡用雙雜湊。
這位大佬寫的可能比較清楚，點選這裡。

程式碼

#include<map>
#include<cstdio>
#include<cstring>
using namespace std;

const int p1 = 19260817;
const int p2 = 23333333;
const int base1 = 89783;
const int base2 = 47881;
int N, M,
 
 ans;
int pow1[200001], pow2[200001], h1[200001], h2[200001];
char s[200001];
map<pair<int, int>, bool> t;

long long find1(int l, int r) {
	return (h1[r] - (long long)h1[l - 1] * pow1[r - l + 1] % p1 + p1) % p1;//取出l到r的子串的雜湊值
}
long long find2(int l,int r) {
	return (h2[r] - (long long)h2[l - 1] * pow2[r - l +
 
 1] % p2 + p2) % p2;//同上
}
int main() {
	scanf("%d %d", &N, &M);
	scanf("%s", s + 1);
	pow1[0] = pow2[0] = 1;
	for (int i = 1; i <= N; i++) {
		pow1[i] = (long long)pow1[i - 1] * base1 % p1;
		pow2[i] = (long long)pow2[i - 1] * base2 % p2;
		h1[i] = ((long long)h1[i - 1] * base1 + s[i] - 96 +
 
 p1) % p1;
		h2[i] = ((long long)h2[i - 1] * base2 + s[i] - 96 + p2) % p2;
	}
	for (int i = M; i <= N; i++) {
		int f1 = find1(i - M + 1, i), f2 = find2(i - M + 1, i);
		if (!t[make_pair(f1, f2)]) {
			t[make_pair(f1, f2)] = 1;
			ans++;
		}
	}
	printf("%d", ans);
}