Spoj SUBST1 New Distinct Substrings

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:
2
CCCCC
ABABA

Output:
5
9

就是让你求一下一个串的本质不同的子串的个数。
这个就等价于求一下后缀自动机每个节点的权值和，每个节点的权值等于(max{}-min{}+1)

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<cstring>
#define ll long long
using namespace std;
#define maxn 200005
int f[maxn],ch[maxn][26],cnt=1;
int n,siz[maxn],l[maxn],T,las=1;
int a[maxn],c[maxn];
char s[maxn];
ll ans=0;

inline void init(){
	cnt=las=1;
	memset(f,0,sizeof(f));
	memset(ch,0,sizeof(ch));
	memset(c,0,sizeof(c));
	siz[1]=l[1]=ans=0;
}

inline void ins(int x){
	int p=las,np=++cnt;
	las=np,l[np]=l[p]+1;
	siz[np]=1;
	
	for(;p&&!ch[p][x];p=f[p]) ch[p][x]=np;
	if(!p) f[np]=1;
	else{
		int q=ch[p][x];
		if(l[q]==l[p]+1) f[np]=q;
		else{
			int nq=++cnt;
			l[nq]=l[p]+1;
			memcpy(ch[nq],ch[q],sizeof(ch[q]));
			f[nq]=f[q];
			f[q]=f[np]=nq;
			for(;ch[p][x]==q;p=f[p]) ch[p][x]=nq;
		}
	}
}

inline void build(){
	for(int i=0;i<n;i++) ins(s[i]-'a');
}

inline void solve(){
	for(int i=1;i<=cnt;i++) ans+=(ll)(l[i]-l[f[i]]);
}

int main(){
	scanf("%d",&T);
	while(T--){
		init();
		scanf("%s",s);
		n=strlen(s);
		build();
		solve();
		printf("%lld
",ans);
	}
	
	return 0;
}