Description
FlyBrother is a superman, therefore he is always busy saving the world.?
To graduate from NUDT is boring but necessary for him. Typically We need to post an paper to get Graduate Certificate, however being one superman, FlyBrother wants to make his paper perfect. A paper is a lower case string. To make it perfect, FlyBrother wanna
the number of different substrings in the paper. It is quite a silly problem for FlyBrother, but because he is so busy, can you help him to solve it?
Input
There are several cases. Process till EOF.
For each case, there is a line of lower case string (1<=length<=100000).
Output
For each case, output one number in a line of the answer described in the problem.
Sample Input
a
aab
Sample Output
1
5
HINT
题意：
求不同的字符串个数
思路：
在我的后缀数组题目小结里有一样的的题目。模板题
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;
#define LS 2*i
#define RS 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 100005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
LL wa[N],wb[N],wsf[N],wv[N],sa[N];
LL rank1[N],height[N],s[N],a[N];
char str[N],str1[N],str2[N];
//sa:字典序中排第i位的起始位置在str中第sa[i]
//rank:就是str第i个位置的后缀是在字典序排第几
//height：字典序排i和i-1的后缀的最长公共前缀
LL cmp(LL *r,LL a,LL b,LL k)
{
return r[a]==r[b]&&r[a+k]==r[b+k];
}
void getsa(LL *r,LL *sa,LL n,LL m)//n要包括末尾加入的0
{
LL i,j,p,*x=wa,*y=wb,*t;
for(i=0; i<m; i++) wsf[i]=0;
for(i=0; i<n; i++) wsf[x[i]=r[i]]++;
for(i=1; i<m; i++) wsf[i]+=wsf[i-1];
for(i=n-1; i>=0; i--) sa[--wsf[x[i]]]=i;
p=1;
j=1;
for(; p<n; j*=2,m=p)
{
for(p=0,i=n-j; i<n; i++) y[p++]=i;
for(i=0; i<n; i++) if(sa[i]>=j) y[p++]=sa[i]-j;
for(i=0; i<n; i++) wv[i]=x[y[i]];
for(i=0; i<m; i++) wsf[i]=0;
for(i=0; i<n; i++) wsf[wv[i]]++;
for(i=1; i<m; i++) wsf[i]+=wsf[i-1];
for(i=n-1; i>=0; i--) sa[--wsf[wv[i]]]=y[i];
t=x;
x=y;
y=t;
x[sa[0]]=0;
for(p=1,i=1; i<n; i++)
x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++;
}
}
void getheight(LL *r,LL n)//n不保存最后的0
{
LL i,j,k=0;
for(i=1; i<=n; i++) rank1[sa[i]]=i;
for(i=0; i<n; i++)
{
if(k)
k--;
else
k=0;
j=sa[rank1[i]-1];
while(r[i+k]==r[j+k])
k++;
height[rank1[i]]=k;
}
}
LL t,ans,n,m;
int main()
{
LL i,j,k,len;
W(~scanf("%s",str))
{
len = strlen(str);
UP(i,0,len-1)
s[i]=str[i];
s[len] = 0;
getsa(s,sa,len+1,300);
getheight(s,len);
ans = (1+len)*len/2;
UP(i,2,len)
ans-=height[i];
printf("%lld\n",ans);
}
}
