Given a string, we need to find the total number of its distinct substrings.

Input

T- number of test cases. T<=20; Each test case consists of one string, whose length is <= 50000

Output

For each test case output one number saying the number of distinct substrings.

Example

Input:

CCCCC

ABABA

Output:

Solution

后缀排序后

一个后缀 \(SA[i]\) 有 \(n-SA[i]+1\) 个子串，但后缀 \(SA[i]\) 与 \(SA[i-1]\) 有 \(height[i]\) 个字符相同，那么就有 \(height[i]\) 个子串一样，减去就是了

#include<bits/stdc++.h>

#define ui unsigned int

#define ll long long

#define db double

#define ld long double

#define ull unsigned long long

const int MAXN=50000+10;

char s[MAXN];

int T,n,m,SA[MAXN],rk[MAXN],cnt[MAXN],nxt[MAXN],height[MAXN];

template<typename T> inline void read(T &x)

{

	T data=0,w=1;

	char ch=0;

	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();

	if(ch=='-')w=-1,ch=getchar();

	while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();

	x=data*w;

}

template<typename T> inline void write(T x,char ch='\0')

{

	if(x<0)putchar('-'),x=-x;

	if(x>9)write(x/10);

	putchar(x%10+'0');

	if(ch!='\0')putchar(ch);

}

template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}

template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}

template<typename T> inline T min(T x,T y){return x<y?x:y;}

template<typename T> inline T max(T x,T y){return x>y?x:y;}

inline void GetSA()

{

	n=strlen(s+1),m=300;

	for(register int i=1;i<=n;++i)rk[i]=s[i];

	for(register int i=1;i<=m;++i)cnt[i]=0;

	for(register int i=1;i<=n;++i)cnt[rk[i]]++;

	for(register int i=1;i<=m;++i)cnt[i]+=cnt[i-1];

	for(register int i=n;i>=1;--i)SA[cnt[rk[i]]--]=i;

	for(register int k=1,ps;k<=n;k<<=1)

	{

		ps=0;

		for(register int i=n-k+1;i<=n;++i)nxt[++ps]=i;

		for(register int i=1;i<=n;++i)

			if(SA[i]>k)nxt[++ps]=SA[i]-k;

		for(register int i=1;i<=m;++i)cnt[i]=0;

		for(register int i=1;i<=n;++i)cnt[rk[i]]++;

		for(register int i=1;i<=m;++i)cnt[i]+=cnt[i-1];

		for(register int i=n;i>=1;--i)SA[cnt[rk[nxt[i]]]--]=nxt[i];

		std::swap(nxt,rk);

		rk[SA[1]]=1;ps=1;

		for(register int i=2;i<=n;rk[SA[i]]=ps,++i)

			if(nxt[SA[i]]!=nxt[SA[i-1]]||nxt[SA[i]+k]!=nxt[SA[i-1]+k])ps++;

		if(ps>=n)break;

		m=ps;

	}

	for(register int i=1,j,k=0;i<=n;height[rk[i++]]=k)

		for(k=k?k-1:k,j=SA[rk[i]-1];s[i+k]==s[j+k];++k);

}

inline int solve()

{

	int ans=0;

	for(register int i=1;i<=n;++i)ans+=n-SA[i]+1-height[i];

	return ans;

}

int main()

{

	read(T);

	while(T--)

	{

		scanf("%s",s+1);

		GetSA();

		write(solve(),'\n');

	}

	return 0;

}

巴特西

【刷题】SPOJ 705 SUBST1 - New Distinct Substrings

Input

Output

Example

Solution

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