POJ 1961 算法中的KMP模式匹配原理是怎样的？
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本文共计522个文字，预计阅读时间需要3分钟。
时间限制：3000MS内存限制：30000K总提交：12709通过：5939
描述：对于给定字符串S的每个前缀，其中S由N个字符组成（每个字符的ASCII码在97到126之间，包括97和126），我们需要知道是否存在一个子串t。
Period
Time Limit:3000MS
Memory Limit:30000K
Total Submissions:12709
Accepted:5939
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 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 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
#include<stdio.h>
#include<string.h>
char a[1000010];
int next[1000010];
int main()
{
int n,m;
int k = 0;
while(scanf("%d",&m)!=EOF && m!=0)
{
scanf("%s",a);
k++;
n = strlen(a);
int i = 0;
int j = -1;
next[i] = -1;
while(i<n)
{
if(j == -1 || a[j] == a[i])
{
i++;
j++;
next[i] = j;
}
else
{
j = next[j];
}
}
printf("Test case #%d\n",k);
for( i=2;i<=n;i++)
{
if(i%(i-next[i])==0)
{
if(i/(i-next[i])>1)
printf("%d %d\n",i,i/(i-next[i]));
}
}
printf("\n");
}
return 0;
}
本文共计522个文字，预计阅读时间需要3分钟。
时间限制：3000MS内存限制：30000K总提交：12709通过：5939
描述：对于给定字符串S的每个前缀，其中S由N个字符组成（每个字符的ASCII码在97到126之间，包括97和126），我们需要知道是否存在一个子串t。
Period
Time Limit:3000MS
Memory Limit:30000K
Total Submissions:12709
Accepted:5939
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 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 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
#include<stdio.h>
#include<string.h>
char a[1000010];
int next[1000010];
int main()
{
int n,m;
int k = 0;
while(scanf("%d",&m)!=EOF && m!=0)
{
scanf("%s",a);
k++;
n = strlen(a);
int i = 0;
int j = -1;
next[i] = -1;
while(i<n)
{
if(j == -1 || a[j] == a[i])
{
i++;
j++;
next[i] = j;
}
else
{
j = next[j];
}
}
printf("Test case #%d\n",k);
for( i=2;i<=n;i++)
{
if(i%(i-next[i])==0)
{
if(i/(i-next[i])>1)
printf("%d %d\n",i,i/(i-next[i]));
}
}
printf("\n");
}
return 0;
}