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poj 2046&&poj1961KMP 前缀数组
Power Strings
Time Limit: 3000 MS Memory Limit: 65536 KB
64-bit integer IO format: %I64d , %I64u Java class name: Main
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Description
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
abcdaaaaababab.
Sample Output
143
题意：求解最多重复子串
利用KMP的前缀数组 以p为模式串 next【i】的意思 为前个字符组成的子串为s 则s的前next【i】个字符与后next【i】个字符相等
注意 ： len（p）-next【len（p））】==循环节的长度
#include <iostream>#include <stdio.h>#include <string.h>using namespace std;int next[1000002];char p[1000002];void find(char p[]){ int m=strlen(p+1); next[1]=0; for(int k=0,q=2; q<=m; q++) { while(k>0&&p[k+1]!=p[q]) k=next[k]; if(p[k+1]==p[q]) k++; next[q]=k; }}int main(){ while(~scanf("%s",p+1)) { if(!strcmp(".",p+1)) break; find(p); int len=strlen(p+1); int len1=len-next[len]; printf("%d\n",len%len1?1:len/len1); }}
Period
Time Limit: 3000 MS Memory Limit: 30000 KB
64-bit integer IO format: %I64d , %I64u Java class name: Main
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Description
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 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.
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
3aaa12aabaabaabaab0
Sample Output
Test case #12 23 3Test case #22 26 29 312 4
题意： 定义字符串A,若A最多由n个相同字串s连接而成，则A=s^n,如"aaa" = "a"^3,"abab" = "ab"^2 "ababa" = "ababa"^1 给出一个字符串A，求该字符串的所有前缀中有多少个前缀SA= s^n（n>1） 输出符合条件的前缀长度及其对应的n
如aaa 前缀aa的长度为2,由2个‘a‘组成 前缀aaa的长度为3，由3个"a"组成
分析：KMP 若某一长度L的前缀符合上诉条件，则 1.next[L]!=0(next[L]=0时字串为原串，不符合条件) 2.L%(L-next[L])==0（此时字串的长度为L/next[L]） 对于2：有str[0]....str[next[L]-1]=str[L-next[L]-1]...str[L-1] =》str[L-next[L]-1] = str[L-next[L]-1+L-next[L]-1] = str[2*(L-next[L]-1)]; 假设S = L-next[L]-1;则有str[0]=str[s]=str[2*s]=str[3*s]...str[k*s],对于所有i%s==0,均有s[i]=s[0] 同理，str[1]=str[s+1]=str[2*s+1].... str[j]=str[s+j]=str[2*s+j].... 综上，若L%S==0,则可得L为str[0]...str[s-1]的相同字串组成， 总长度为L,其中字串长度SL = s-0+1=L-next[L],循环次数为L/SL 故对于所有大于1的前缀，只要其符合上述条件，即为答案之一#include "stdio.h"int p[1000010],N;char str[1000010];void get_p(int n){ int i,j=-1; p[0]=-1; for(i=1;i<n;i++) { while(j>-1 && str[i]!=str[j+1]) j=p[j]; if(str[i] == str[j+1]) j++; p[i]=j; }}int main(){ int i,j,cas=1; while(scanf("%d",&N),N) { scanf("%s",str); get_p(N); printf("Test case #%d\n",cas++); for(i=1;i<N;i++) { if(p[i]!=-1 && (i+1)%(i-p[i])==0) printf("%d %d\n",i+1,(i+1)/(i-p[i])); } printf("\n"); }}
http://www.cnblogs.com/dolphin0520/archive/2011/08/24/2151846.html 看一看
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