3026 - Period
Time limit: 3.000 seconds
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 AK , 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
本题用了MP算法
显然i能和f[i]匹配，那么字符串刚好可以连着等下去
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
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<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;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]; //字符串从0开始，但是f[i]表示匹配第i个字符，前面留一个 f[0]--a-->f[1]--...这样的
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;
}