LA 3026(Period

3026 - Period

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 ≤ i ≤ N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A^K , that is A concatenated K times, for some string A. Of course, we also want to know the period K.

Input

The input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 ≤ N ≤ 1 000 000) the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.

Output

For each test case, output 'Test case #' and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

Sample Input

3
aaa
12
aabaabaabaab
0

Sample Output

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

本題用了MP算法

顯然i能和f[i]匹配，那麼字符串剛好可以連著等下去

<喎?/kf/ware/vc/" target="_blank" class="keylink">vcD4KPHA+PGJyPgo8L3A+CjxwPjxicj4KPC9wPgo8cD48cHJlIGNsYXNzPQ=="brush:java;">#include #include #include #include #include #include #include #include #include using namespace std; #define For(i,n) for(int i=1;i<=n;i++) #define Fork(i,k,n) for(int i=k;i<=n;i++) #define Rep(i,n) for(int i=0;i=0;i--) #define Forp(x) for(int p=pre[x];p;p=next[p]) #define Forpiter(x) for(int &p=iter[x];p;p=next[p]) #define Lson (x<<1) #define Rson ((x<<1)+1) #define MEM(a) memset(a,0,sizeof(a)); #define MEMI(a) memset(a,127,sizeof(a)); #define MEMi(a) memset(a,128,sizeof(a)); #define INF (2139062143) #define F (100000007) #define MAXN (1000000+10) typedef long long ll; ll mul(ll a,ll b){return (a*b)%F;} ll add(ll a,ll b){return (a+b)%F;} ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;} void upd(ll &a,ll b){a=(a%F+b%F)%F;} // kmp class KMP { public: int f2[MAXN]; //字符串從0開始，但是f[i]表示匹配第i個字符，前面留一個 f[0]--a-->f[1]--...這樣的 char T2[MAXN],P2[MAXN]; //T is long,P is model str void mem(){MEM(f2) MEM(T2) MEM(P2) } int getFail(char *P=0,int* f=0) { if (P==0) P=P2;if (f==0) f=f2; int m=strlen(P); f[0]=f[1]=0; For(i,m-1) { int j=f[i]; while(j&&P[i]!=P[j]) j=f[j]; f[i+1]= P[i] == P[j] ? j+1 : 0; } } int find(char* T=0,char* P=0,int* f=0) { if (T==0) T=T2;if (P==0) P=P2;if (f==0) f=f2; int n=strlen(T),m=strlen(P); getFail(P,f); int j=0; Rep(i,n) { while(j&&T[i]!=P[j]) j=f[j]; if (T[i]==P[j]) j++; if (j==m) return i-m+1; } } }S; int n; int main() { // freopen("la3026.in","r",stdin); // freopen(".out","w",stdout); int tt=0; while(scanf("%d%s",&n,S.P2)==2) { printf("Test case #%d\n",++tt); S.getFail(S.P2,S.f2); Fork(i,2,n) { if (i%(i-S.f2[i])==0&&S.f2[i]) printf("%d %d\n",i,i/(i-S.f2[i])); } putchar('\n'); S.mem(); } return 0; }