hdu 1358:Period（KMP算法，next[]数组的使用）

Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2398 Accepted Submission(s): 1187


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

3

aaa

12

aabaabaabaab

0

[align=left]Sample Output[/align]

Test case #1

2 2

3 3

Test case #2

2 2

6 2

9 3

12 4

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　　KMP算法。
　　这道题考察的是KMP算法中next数组的应用，必须理解透next[]数组代表的含义才能通过它解决这道题。
　　思路是先构造出 next[] 数组，下标为 i，定义一个变量 j = i - next[i] 就是next数组下标和下标对应值的差，如果这个差能整除下标 i，即 i%j==0 ,则说明下标i之前的字符串（周期性字符串长度为 i）一定可以由一个前缀周期性的表示出来，这个前缀的长度为刚才求得的那个差，即 j，则这个前缀出现的次数为 i/j 。所以最后输出i和i/j即可。

　　举这道题的第二组输入样例为例：
　　其next[]数组为：

i      0  1  2  3  4  5  6  7  8  9  10 11
a[i]    a  a  b  a  a  b  a  a  b  a  a  b
next[i] -1  0  1  0  1  2  3  4  5  6  7  8

　　　　　　　　　　　　 [b]↓[/b]



　　next[i]值是0或-1的忽略。

　　注意：由于输出次数太多 (2 <= N <= 1 000 000)，建议用printf输出，否则会超时。

　　代码：

#include <iostream>
#include <stdio.h>
using namespace std;
char a[1000010];
int next[1000010];
int n;
void GetNext()    //获得a数列的next数组
{
int i=0,k=-1;
next[0] = -1;
while(i<n){
if(k==-1){
next[i+1] = 0;
i++;k++;
}
else if(a[i]==a[k]){
next[i+1] = k+1;
i++;k++;
}
else
k = next[k];
}
}
void DisRes(int num)
{
int j;
printf("Test case #%d\n",num);
for(int i=0;i<=n;i++){
if(next[i]==-1 || next[i]==0)   //next[i]是-1或0的忽略，说明之前没有周期性前缀
continue;
j = i - next[i];
if(i%j==0)  //能整除，说明存在周期性前缀
printf("%d %d\n",i,i/j);    //输出这个前缀的长度和周期数
}
printf("\n");
}
int main()
{
int num = 0;
while(scanf("%d",&n)!=EOF){
if(n==0) break;
scanf("%s",a);
GetNext();  //获得next[]数组
DisRes(++num);  //输出结果
}
return 0;
}

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