pat 1005 Programming Pattern (35 分)

Programmers often have a preference among program constructs. For example, some may prefer if(0==a), while others may prefer if(!a). Analyzing such patterns can help to narrow down a programmer's identity, which is useful for detecting plagiarism.

Now given some text sampled from someone's program, can you find the person's most commonly used pattern of a specific length?

Input Specification:

Each input file contains one test case. For each case, there is one line consisting of the pattern length

Output Specification:

For each test case, print in one line the length-

Whitespace characters in the input should be printed as they are. Also note that there may be multiple occurrences of the same substring overlapping each other.

Sample Input 1:

4
//A can can can a can.

Sample Output 1:

 can 4

Sample Input 2:

3
int a=~~~~~~~~~~~~~~~~~~~~~0;

Sample Output 2:

~~~ 19

这题是字符串很常规的题目，可以用后缀数组来做，但是我是用hash硬搞的，使用了自然溢出，我甚至连字典序都没盘，就过了，再次说明pat的数据实在是太水了。  
另外，pat的g++编译器不能用gets是什么鬼，求g++的版本？？？

 1 #include<cstdio> 
 2 #include<cstring>  
 3 #include<cmath>
 4 #include<iostream>
 5 #include<algorithm>  
 6 #include<cstdlib> 
 7 using namespace std;  
 8 #define ui  unsigned long long  
 9 int const N=1048576+100;  
10 ui  const p=100007;   
11 int n;  
12 ui h[N],pw[N];  
13 char s[N],ans[N];   
14 struct node{
15     ui v;  
16     int id;  
17 }a[N];  
18 int cmp(node x,node y){
19     return x.v<y.v;  
20 }
21 int main(){
22     scanf("%d",&n); 
23     getchar();  
24     scanf("%[^\n]",s+1);  
25     int len=strlen(s+1);  
26     pw[0]=1;   
27     for(int i=1;i<=len;i++)  
28         pw[i]=pw[i-1]*p;  
29     for(int i=1;i<=n;i++)  
30         h[n]=h[n]*p+s[i]; 
31     for(int i=n+1;i<=len;i++){
32         h[i]=(h[i-1]-s[i-n]*pw[n-1])*p+s[i];  
33     }
34     for(int i=n;i<=len;i++){
35         a[i].id=i;  
36         a[i].v=h[i];  
37     }
38     sort(a+n,a+len+1,cmp);  
39     int sum=0,cnt=0,x;  
40     for(int i=n;i<=len;i++) 
41     {
42         if(a[i].v==a[i-1].v) sum++;  
43         else sum=1;     
44         if(sum>cnt){
45             x=a[i].id;cnt=sum;  
46         } 
47     } 
48     for(int i=x-n+1;i<=x;i++)  
49         printf("%c",s[i]);  
50     printf(" ");  
51     printf("%d\n",cnt);  
52     return 0; 
53 }

View Code

后缀数组

 1 #include<bits/stdc++.h>
 2 using namespace std;  
 3 int const N=(1<<20)+10;  
 4 int wa[N<<1],wb[N<<1],num[N],n,rk[N],sa[N],h[N],wv[N];  
 5 char s[N];  
 6 int inline cmp(int *r,int x,int y,int z){
 7   return r[x]==r[y] && r[x+z]==r[y+z];  
 8 }
 9 void build_sa(char *r,int n,int m){
10   int *x=wa,*y=wb;  
11   for(int i=0;i<m;i++)  num[i]=0;  
12   for(int i=0;i<n;i++)  num[x[i]=r[i]]++; 
13   for(int i=1;i<m;i++)  num[i]+=num[i-1]; 
14   for(int i=n-1;i>=0;i--)  sa[--num[x[i]]]=i; 
15   for(int j=1,p=0;p<n;j<<=1,m=p){
16     p=0;  
17     for(int i=n-j;i<n;i++) y[p++]=i; 
18     for(int i=0;i<n;i++) if(sa[i]>=j)  y[p++]=sa[i]-j;  
19     for(int i=0;i<m;i++) num[i]=0;  
20     for(int i=0;i<n;i++) num[wv[i]=x[y[i]]]++;  
21     for(int i=1;i<m;i++) num[i]+=num[i-1];  
22     for(int i=n-1;i>=0;i--) sa[--num[wv[i]]]=y[i];  
23     swap(x,y); 
24     p=1;x[sa[0]]=0;  
25     for(int i=1;i<n;i++)  
26       x[sa[i]]=cmp(y,sa[i],sa[i-1],j)? p-1:p++;  
27   }  
28   for(int i=0;i<n;i++) rk[i]=x[i];  
29 }
30 void build_h(char *r,int n){
31   int k=0;  
32   for(int i=0;i<n;i++){
33     if(k) k--;  
34     int j=sa[rk[i]-1];  
35     while (r[i+k]==r[j+k]) k++;  
36     h[rk[i]]=k;  
37   }
38 }
39 void solve(int len){
40   int ans=0,id,cnt=0;    
41   for(int i=1;i<=len;i++){
42     if(h[i]>=n){
43       if(cnt==0) cnt=2;  
44       else cnt++;  
45     }else cnt=0; 
46     if(cnt>ans) {
47       ans=cnt;  
48       id=sa[i];   
49     }
50   }
51   if(ans==0) id=sa[1],ans=1;  
52   for(int i=id;i<=id+n-1;i++)  
53     putchar(s[i]); 
54   putchar(' ');  
55   printf("%d\n",ans);  
56 }
57 int main(){
58   scanf("%d",&n); 
59   getchar();  
60   scanf("%[^\n]",s);  
61   int len=strlen(s);  
62   build_sa(s,len+1,130);    
63   build_h(s,len);   
64   solve(len);  
65   return 0;  
66 }