大意

判斷在一個長度為$n$的串中長度為$m$的不同連續子串個數

思路

裸的$hash$，然而被卡了，所以要用 雙模數

程式碼

#include<cstdio>
#include<algorithm>
#define ymw 1000000007
#define bpm 1000000009//雙模數
using namespace std;typedef unsigned long long ull;
int n,m,len,ans=1;
ull h[200001],c[200001];
char s[200001];
struct node
{
	int x,y;
	bool operator<(node h)
 
const
	{
		return x==h.x?y<h.y:x<h.x;
    }
}a[200001];
signed main()
{
	scanf("%d%d\n",&n,&m);
	gets(s+1);
	c[0]=1;
	for(register int i=1;i<=n;i++)
	{
		h[i]=(h[i-1]*437+s[i]-96)%ymw;
		c[i]=c[i-1]*437%ymw;
	}
	for(register int i=1;i+m-1<=n;i++) a[i].x=(h[i+m-1]-h[i-1]*c[m]%ymw+ymw)%
 
ymw;
	c[0]=1;h[0]=0;
	for(register int i=1;i<=n;i++)
	{
		h[i]=(h[i-1]*439+s[i]-96)%bpm;
		c[i]=c[i-1]*439%bpm;
	}
	for(register int i=1;i+m-1<=n;i++) a[i].y=(h[i+m-1]-h[i-1]*c[m]%bpm+bpm)%bpm;
	sort(a+1,a+1+n-m+1);
	for(register int i=2;i+m-1<=n;i++) ans+=(a[i].x!=a[i-1].x||a[i].y!=a[i-1].y);
 
//離散化
	printf("%d",ans);//輸出
}