单哈希且用自然溢出代替取模操作，常数小但是容易被卡

单字符串区间内比较

typedef unsigned long long ULL;

const int N = 100010, P = 131;

int n, m;
char str[N];
ULL h[N], p[N];

ULL get(int l, int r)
{
    return h[r] - h[l - 1] * p[r - l + 1];
}

int main()
{
    scanf("%d%d", &n, &m);
    scanf("%s", str + 1);

    p[0] = 1;
    for (int i = 1; i <= n; i ++ )
    {
        h[i] = h[i - 1] * P + str[i];
        p[i] = p[i - 1] * P;
    }

    while (m -- )
    {
        int l1, r1, l2, r2;
        scanf("%d%d%d%d", &l1, &r1, &l2, &r2);

        if (get(l1, r1) == get(l2, r2)) puts("Yes");
        else puts("No");
    }

    return 0;
}

# 字符串双哈希，多字符串互相比较
const int mod = 998244353;
const ull mod1=212370440130137957ll;
const ull mod2=(1<<30);
const int base=233333;
const double eps = 1e-8;
int n, m;



struct node{ull x,y;
    bool operator < (const node& a)const{
        return a.x==x?a.y<y:a.x<x;
    }
}f[N];
ull hash1(string s){
	const int base1=1313131;
    ull ans=0,len=s.size();
    for(int i=0;i<len;++i)ans=(ans*base1+(ull)s[i])%mod1;
    return ans;
}
ull hash2(string s){
    ull ans=0,len=s.size();
    const int base2=233333;
    for(int i=0;i<len;++i)ans=(ans*base2+(ull)s[i])%mod2;
    return ans;
}
void solve(){
	cin>>n;
	for(int i=1;i<=n;i++){
		string s;
		cin>>s;
		f[i].x=hash1(s);
		f[i].y=hash2(s);
	}
	sort(f+1,f+n+1);
	int ans=0;
	for(int i=1;i<=n;i++){
		if(f[i].x==f[i+1].x&&f[i].y==f[i+1].y)continue;
		else ans++;
	}
	cout<<ans<<endl;
	
}