Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).
Input
Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.
Output
For each s you should print the largest n such that s = a^n for some string a.
Sample Input
abcd aaaa ababab .
Sample Output
1 4 3
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
题意:
输入一个字符串，让你找出来这个串的最小循环节
题解：
对于答案就用for循环从1到n枚举（为什么不能用二分？因为这个判断没有单调性。即i不是循环节的话，你不能说i-1或者i+1也肯定不是循环节）
下一步就是判断枚举的这个i是不是循环节，怎么判断？
满足这三个条件：n%i==0 && Rank[0]-Rank[i]==1 && height[Rank[0]]==n-i
1、n%i==0这就肯定不需要说了
2、你要注意Rank[0]表示的是啥，他代表后缀[0...n]的排名是多少，如果循环节是i，那么Rank[0]-Rank[i]==Rank[i]-Rank[2*i]......==1
3、如果循环节是i，那么height[Rank[0]]==n-i, height[Rank[0]]代表后缀[0..n]与后缀[i...n]的最长公共前缀，既然循环节是i，那么height[Rank[0]]的值也就等于n-i
还要注意这道题用后缀数组来做的话要用dc3模板，如果用Da模板会TLE
dc3模板AC代码：
1 #include <cstdlib> 2 #include <cstring> 3 #include <cstdio> 4 #include <algorithm> 5 using namespace std; 6 const int maxn = 3000010; 7 #define F(x) ((x) / 3 + ((x) % 3 == 1 ? 0 : tb)) 8 #define G(x) ((x) < tb ? (x) * 3 + 1 : ((x) - tb) * 3 + 2) 9 int wa[maxn], wb[maxn], Ws[maxn], wv[maxn], sa[maxn]; 10 int Rank[maxn], height[maxn],r[maxn]; 11 char s[maxn]; 12 int c0(int *r, int a, int b) 13 { 14 return r[a] == r[b] && r[a + 1] == r[b + 1] && r[a + 2] == r[b + 2]; 15 } 16 int c12(int k, int *r, int a, int b) 17 { 18 if (k == 2) 19 return r[a] < r[b] || r[a] == r[b] && c12(1, r, a + 1, b + 1); 20 return r[a] < r[b] || r[a] == r[b] && wv[a + 1] < wv[b + 1]; 21 } 22 void Rsort(int *r, int *a, int *b, int n, int m) 23 { 24 for (int i = 0; i < n; i++) wv[i] = r[a[i]]; 25 for (int i = 0; i < m; i++) Ws[i] = 0; 26 for (int i = 0; i < n; i++) Ws[wv[i]]++; 27 for (int i = 1; i < m; i++) Ws[i] += Ws[i - 1]; 28 for (int i = n - 1; i >= 0; i--) b[--Ws[wv[i]]] = a[i]; 29 } 30 void dc3(int *r,int *sa,int n, int m) 31 { 32 int i, j, *rn = r + n, *san = sa + n, ta = 0, tb = (n + 1) / 3, tbc = 0, p; 33 r[n] = r[n + 1] = 0; 34 for (i = 0; i < n; i++) if (i % 3 != 0) wa[tbc++] = i; 35 Rsort(r + 2, wa, wb, tbc, m); 36 Rsort(r + 1, wb, wa, tbc, m); 37 Rsort(r, wa, wb, tbc, m); 38 for (p = 1, rn[F(wb[0])] = 0, i = 1; i < tbc; i++) 39 rn[F(wb[i])] = c0(r, wb[i - 1], wb[i]) ? p - 1 : p++; 40 if (p < tbc) dc3(rn, san, tbc, p); 41 else for (i = 0; i < tbc; i++) san[rn[i]] = i; 42 for (i = 0; i < tbc; i++) if (san[i] < tb) wb[ta++] = san[i] * 3; 43 if (n % 3 == 1) wb[ta++] = n - 1; 44 Rsort(r, wb, wa, ta, m); 45 for (i = 0; i < tbc; i++) wv[wb[i] = G(san[i])] = i; 46 for (i = 0, j = 0, p = 0; i < ta && j < tbc; p++) 47 sa[p] = c12(wb[j] % 3, r, wa[i], wb[j]) ? wa[i++] : wb[j++]; 48 for (; i < ta; p++) sa[p] = wa[i++]; 49 for (; j < tbc; p++) sa[p] = wb[j++]; 50 } 51 void get_height(int n) 52 { 53 int i, j, k = 0; 54 for (i = 1; i <= n; i++) Rank[sa[i]] = i; 55 for (i = 0; i < n; height[Rank[i++]] = k) 56 for (k ? k-- : 0, j = sa[Rank[i] - 1]; r[i + k] == r[j + k]; k++); 57 } 58 int main() 59 { 60 while(~scanf("%s",s)) 61 { 62 int n=strlen(s); 63 if(n==1 && s[0]=='.') break; 64 for(int i=0;i<n;++i) 65 r[i]=s[i]; 66 r[n]='0'; 67 dc3(r,sa,n+1,200); 68 get_height(n); 69 int flag=0; 70 for(int i=1;i<n;++i) //枚举循环节长度 71 { 72 if(n%i==0 && Rank[0]-Rank[i]==1 && height[Rank[0]]==n-i) 73 { 74 flag=1; 75 printf("%d ",n/i); 76 break; 77 } 78 } 79 if(!flag) 80 printf("1 "); 81 } 82 return 0; 83 }
Da模板TLE代码：
1 #include <cstdlib> 2 #include <cstring> 3 #include <cstdio> 4 #include <algorithm> 5 using namespace std; 6 7 const int N = 1000010; 8 int x[N], y[N], c[N]; 9 int rank[N], height[N]; 10 int sa[N],n,k; 11 char s[N]; 12 bool pan(int *x,int i,int j,int k,int n) 13 { 14 int ti=i+k<n?x[i+k]:-1; 15 int tj=j+k<n?x[j+k]:-1; 16 return x[i]==x[j]&&ti==tj; 17 } 18 void build_SA(int n,int r) 19 { 20 int *x=rank,*y=height; 21 for(int i=0; i<r; i++)c[i]=0; 22 for(int i=0; i<n; i++)c[s[i]]++; 23 for(int i=1; i<r; i++)c[i]+=c[i-1]; 24 for(int i=n-1; i>=0; i--)sa[--c[s[i]]]=i; 25 r=1; 26 x[sa[0]]=0; 27 for(int i=1; i<n; i++) 28 x[sa[i]]=s[sa[i]]==s[sa[i-1]]?r-1:r++; 29 for(int k=1; r<n; k<<=1) 30 { 31 int yn=0; 32 for(int i=n-k; i<n; i++)y[yn++]=i; 33 for(int i=0; i<n; i++) 34 if(sa[i]>=k)y[yn++]=sa[i]-k; 35 for(int i=0; i<r; i++)c[i]=0; 36 for(int i=0; i<n; i++)++c[x[y[i]]]; 37 for(int i=1; i<r; i++)c[i]+=c[i-1]; 38 for(int i=n-1; i>=0; i--)sa[--c[x[y[i]]]]=y[i]; 39 swap(x,y); 40 r=1; 41 x[sa[0]]=0; 42 for(int i=1; i<n; i++) 43 x[sa[i]]=pan(y,sa[i],sa[i-1],k,n)?r-1:r++; 44 } 45 for(int i=0; i<n; i++)rank[i]=x[i]; 46 } 47 void get_height(int n) 48 { 49 int i,j,k=0; 50 for(i=1; i<=n; i++)rank[sa[i]]=i; 51 for(i=0; i<n; i++) 52 { 53 if(k)k--; 54 else k=0; 55 j=sa[rank[i]-1]; 56 while(s[i+k]==s[j+k])k++; 57 height[rank[i]]=k; 58 } 59 } 60 int check(int len) 61 { 62 int i=2,cnt=0; 63 while(1) 64 { 65 while(i<=n && height[i]>=len) 66 cnt++,i++; 67 if(cnt+1>=k)return 1; 68 if(i>=n)return 0; 69 while(i <=n &&height[i]<len) 70 i++; 71 cnt=0; 72 } 73 } 74 75 int main() 76 { 77 while(~scanf("%s",s)) 78 { 79 n=strlen(s); 80 if(n==1 && s[0]=='.') break; 81 s[n]='0'; 82 build_SA(n+1,200); 83 get_height(n); 84 //printf("%d %d %d ",rank[0],rank[3],height[rank[0]]); 85 int flag=0; 86 for(int i=1;i<n;++i) //枚举循环节长度 87 { 88 if(n%i==0 && rank[0]-rank[i]==1 && height[rank[0]]==n-i) 89 { 90 flag=1; 91 printf("%d ",n/i); 92 break; 93 } 94 } 95 if(!flag) 96 printf("1 "); 97 } 98 return 0; 99 }