poj 1961 Period(kmp最短循环节)
【摘要】
题目链接
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 whet...
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.……………………
题意:
给定一个长度为n的字符串s,求他每个前缀的最短循环节。换句话说,对于每个i(2<=i<=n),求一个最大的整数k(如果k存在),使得s的前i个字符可以组成的前缀是某个字符串重复k次得到的。输出所有存在K的i和对应的k。
这是刘汝佳《算法竞赛入门经典训练指南》上的原题(p213),用KMP构造状态转移表。
解题代码(完全参照白书):
//poj 1961
#include <stdio.h>
const int maxn = 1000000 + 10;
char p[maxn];
int f[maxn];
int main()
{
int n, t = 1;
while (scanf("%d",&n) && n)
{
scanf("%s",p);
f[0] = 0;
f[1] = 0;
for (int i = 1; i < n; i++)
{
int j = f[i];
while (p[i] != p[j] && j)
j = f[j];
f[i+1] = (p[i] == p[j] ? j+1 : 0);
}/*这段代码就是KMP里面核心的代码,这里没有文本串,
我们只需要处理模板即可*/
printf("Test case #%d\n",t++);
for (int i = 2; i <= n; i++)
{
if (f[i] > 0 && i % (i-f[i]) == 0)
printf("%d %d\n",i, i / (i-f[i]));
}
puts("");
}
return 0;
}
文章来源: xindoo.blog.csdn.net,作者:xindoo,版权归原作者所有,如需转载,请联系作者。
原文链接:xindoo.blog.csdn.net/article/details/8764044
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