设为首页 加入收藏

TOP

CSU1656: Paper of FlyBrother(后缀数组)
2015-11-21 00:59:11 来源: 作者: 【 】 浏览:2
Tags:CSU1656: Paper FlyBrother 后缀

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 
  
   
#include 
   
     #include 
    
      #include 
     
       #include 
      
        #include
        #include 
        
          #include 
         
           #include 
          
            #include 
           
             #include 
            
              #include 
             
               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-8www.2cto.com 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要包含末尾添加的0 { LL i,j,p,*x=wa,*y=wb,*t; for(i=0; i
              
               =0; i--) sa[--wsf[x[i]]]=i; p=1; j=1; for(; p
               
                =j) y[p++]=sa[i]-j; for(i=0; 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
                  
                 
                
               
              
             
            
           
          
         
        
      
     
    
   
  
】【打印繁体】【投稿】【收藏】 【推荐】【举报】【评论】 【关闭】 【返回顶部
分享到: 
上一篇CSU1664: 防水堤坝 下一篇hdu 2952 Counting Sheep

评论

帐  号: 密码: (新用户注册)
验 证 码:
表  情:
内  容: