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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#292291#5666. Repetitive ElementsCrysflyAC ✓1ms5236kbC++142.6kb2023-12-27 22:35:432023-12-27 22:35:43

Judging History

你现在查看的是最新测评结果

  • [2023-12-27 22:35:43]
  • 评测
  • 测评结果:AC
  • 用时:1ms
  • 内存:5236kb
  • [2023-12-27 22:35:43]
  • 提交

answer

#include<bits/stdc++.h>
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
#define ll long long
#define ull unsigned long long
using namespace std;
inline ll read()
{
	char c=getchar();ll x=0;bool f=0;
	for(;!isdigit(c);c=getchar())f^=!(c^45);
	for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+(c^48);
	if(f)x=-x;return x;
}
// modint
#define mod 998244353
struct modint{
	int x;
	modint(int o=0){x=o;}
	modint &operator = (int o){return x=o,*this;}
	modint &operator +=(modint o){return x=x+o.x>=mod?x+o.x-mod:x+o.x,*this;}
	modint &operator -=(modint o){return x=x-o.x<0?x-o.x+mod:x-o.x,*this;}
	modint &operator *=(modint o){return x=1ll*x*o.x%mod,*this;}
	modint &operator ^=(int b){
		modint a=*this,c=1;
		for(;b;b>>=1,a*=a)if(b&1)c*=a;
		return x=c.x,*this;
	}
	modint &operator /=(modint o){return *this *=o^=mod-2;}
	friend modint operator +(modint a,modint b){return a+=b;}
	friend modint operator -(modint a,modint b){return a-=b;}
	friend modint operator *(modint a,modint b){return a*=b;}
	friend modint operator /(modint a,modint b){return a/=b;}
	friend modint operator ^(modint a,int b){return a^=b;}
	friend bool operator ==(modint a,int b){return a.x==b;}
	friend bool operator !=(modint a,int b){return a.x!=b;}
	bool operator ! () {return !x;}
	modint operator - () {return x?mod-x:0;}
	bool operator <(const modint&b)const{return x<b.x;}
};
inline modint qpow(modint x,int y){return x^y;}

vector<modint> fac,ifac,iv;
inline void initC(int n)
{
	if(iv.empty())fac=ifac=iv=vector<modint>(2,1);
	int m=iv.size(); ++n;
	if(m>=n)return;
	iv.resize(n),fac.resize(n),ifac.resize(n);
	For(i,m,n-1){
		iv[i]=iv[mod%i]*(mod-mod/i);
		fac[i]=fac[i-1]*i,ifac[i]=ifac[i-1]*iv[i];
	}
}
inline modint C(int n,int m){
	if(m<0||n<m)return 0;
	return initC(n),fac[n]*ifac[m]*ifac[n-m];
}
inline modint sign(int n){return (n&1)?(mod-1):(1);}

#define fi first
#define se second
#define pb push_back
#define mkp make_pair 
typedef pair<int,int>pii;
typedef vector<int>vi;

#define maxn 200005
#define inf 0x3f3f3f3f

int n;
char s[maxn];
modint h[maxn],pw[maxn];

modint H(int l,int r){
	return h[r]-h[l-1]*pw[r-l+1];
}

void work()
{
	cin>>(s+1);
	n=strlen(s+1);
	For(i,1,n)h[i]=h[i-1]*233+s[i];
	pii res; int mx=0;
	For(i,1,n){
		For(j,i+1,n)
			if(s[i]==s[j]){
				int k=1;
				while(j+k<=n && i+k<j && s[i+k]==s[j+k])++k;
				if(k>mx)mx=k,res=mkp(i,i+k-1);
			}
	}
	For(i,res.first,res.se) cout<<s[i]; puts("");
}

signed main()
{
	pw[0]=1;
	For(i,1,10000)pw[i]=pw[i-1]*233;
	int T=read();
	while(T--)work();
	return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 5236kb

input:

5
TATCGATCGAGTTGT
TCCGCGAGCGAGTCTCTCCATT
GTTTCATCATACGAGGCCCCATACGCGCTGG
AGATGGGATCCTTATG
GCCCTTAGGCATGGGATGTCGTTTCTTG

output:

ATCG
GCGA
CATACG
GAT
CTT

result:

ok 5 lines

Test #2:

score: 0
Accepted
time: 1ms
memory: 5168kb

input:

50
TTGACAACTTCAGGTTGGCACTCCTTCATTTGGATTTCGGAATAATAGTTTTCTGCTCTGCC
ATCCTATTCGGGGATAGGAGAGATGGGTTGCCGCTATAAAAGCATTTGAACTCCATTTCACTCCGTTGGCTAGGGGTCGCACTG
CCGTAATATAAAGACTCGGAATTCCAATAGCTGCTATTTGCGAGTATGTGACTGAAAACACACCTATAAATATTAGCTGCGTACAAGCTA
ATGGCTGCATGCAGGGTCGACTAGACACACTTTGTCT
TTGAGGATGTCGACGTGTCT...

output:

CTTCA
CATTT
TAGCTGC
TGCA
ACGTG
GCGCCGG
CTCTT
AGTAT
AGAG
ACAG
TAT
TGAC
CTTG
CGTC
TACTGG
GCCGGT
GAA
CAGTA
GCGT
GGTT
CCCT
GAG
TAGAC
GGTGC
GCAGT
TGAG
ATCAA
CCACACA
GAGTC
ATGTA
ATGGTA
TATA
TATGAA
TTCC
CATACG
TACCA
TTAG
GGAATGT
CAGG
GCT
AAG
CTGT
GGAT
TCTTC
AAAAC
ATG
GATAA
TTA
ACATAT
CAAT

result:

ok 50 lines