题目链接

P10467 [CCC 2007] Snowflake Snow Snowflakes

解题思路

字符串哈希板子题。

思路就是我们给每个数列的所有排列都哈希一个值，然后判断是否有不同的数列的哈希值相同，如果有，就输出 Twin snowflakes found.，否则就输出 No two snowflakes are alike.。

参考代码

这里使用双哈希。

#include<bits/stdc++.h>
using namespace std;
#define map unordered_map
#define forl(i,a,b) for(register long long i=a;i<=b;i++)
#define forr(i,a,b) for(register long long i=a;i>=b;i--)
#define forll(i,a,b,c) for(register long long i=a;i<=b;i+=c)
#define forrr(i,a,b,c) for(register long long i=a;i>=b;i-=c)
#define lc(x) x<<1
#define rc(x) x<<1|1
//#define mid ((l+r)>>1)
#define cin(x) scanf("%lld",&x)
#define cout(x) printf("%lld",x)
#define lowbit(x) (x&-x)
#define pb push_back
#define pf push_front
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define endl '\n'
#define QwQ return 0;
#define ll long long
#define ull unsigned long long
#define lcm(x,y) x/__gcd(x,y)*y
#define Sum(x,y) 1ll*(x+y)*(y-x+1)/2
#define aty cout<<"Yes\n";
#define atn cout<<"No\n";
#define cfy cout<<"YES\n";
#define cfn cout<<"NO\n";
#define xxy cout<<"yes\n";
#define xxn cout<<"no\n";
#define printcf(x) x?cout<<"YES\n":cout<<"NO\n";
#define printat(x) x?cout<<"Yes\n":cout<<"No\n";
#define printxx(x) x?cout<<"yes\n":cout<<"no\n";
ll t;
string s;
ll q,l1,l2,r1,r2;
long long Ss=chrono::steady_clock::now().time_since_epoch().count();
mt19937_64 Apple(Ss);
long long rand_lr(ll l,ll r){
	return Apple()%(r-l+1)+l;
}
ll pw1[1000010],pw2[1000010],pw3[1000010];
ll hash1[1000010],hash2[1000010],hash3[1000010];
ll Base1=779,Base2=223,Base3=667;
ll mod1=1e9+7,mod2=1e9+9,mod3=998244353;
void init()
{
	pw1[0]=pw2[0]=pw3[0]=1;
	forl(i,1,1e6+5)
		pw1[i]=pw1[i-1]*Base1;
	forl(i,1,1e6+5)
		pw2[i]=pw2[i-1]*Base2;
	forl(i,1,1e6+5)
		pw3[i]=pw3[i-1]*Base3;
	forl(i,1,(ll)s.size()-1)
		hash1[i]=(hash1[i-1]*Base1+s[i]);
	forl(i,1,(ll)s.size()-1)
		hash2[i]=(hash2[i-1]*Base2+s[i]);
	forl(i,1,(ll)s.size()-1)
		hash3[i]=(hash3[i-1]*Base3+s[i]);		
}
void solve()
{
	cin>>s>>q;
	s=' '+s;
	init();
	while(q--)
	{
		ll l1,l2,r1,r2;
		cin>>l1>>r1>>l2>>r2;
		printat((hash1[r1]-hash1[l1-1]*pw1[r1-l1+1])==(hash1[r2]-hash1[l2-1]*pw1[r2-l2+1]) &&
				(hash2[r1]-hash2[l1-1]*pw2[r1-l1+1])==(hash2[r2]-hash2[l2-1]*pw2[r2-l2+1]) &&
				(hash3[r1]-hash3[l1-1]*pw3[r1-l1+1])==(hash3[r2]-hash3[l2-1]*pw3[r2-l2+1]) );
	}
}
int main()
{
	IOS;
	t=1;
//	cin>>t;
	while(t--)
		solve();
	QwQ;
}