题目链接:https://vjudge.net/problem/POJ-1961
题意:给定一个长为n的字符串(n<=1e6),对于下标i(2<=i<=n),如果子串s(1...i)是周期子串,输出其最大周期。
思路:
考察对kmp算法中next数组的定义掌握,如果(i+1)%(i-j)==0 && (i+1)/(i-j) > 1,那么该子串即为满足条件。
AC代码:
#include<cstdio> #include<algorithm> #include<cstring> using namespace std; const int maxn=1e6+5; int n,cas,nex[maxn]; char s[maxn]; void get_next(){ int j; j=nex[0]=-1; for(int i=1;i<n;++i){ while(j>-1&&s[i]!=s[j+1]) j=nex[j]; if(s[i]==s[j+1]) ++j; nex[i]=j; if((i+1)%(i-j)==0&&(i+1)/(i-j)>1) printf("%d %d\n",i+1,(i+1)/(i-j)); } } int main(){ while(scanf("%d",&n),n){ scanf("%s",s); printf("Test case #%d\n",++cas); get_next(); printf("\n"); } return 0; }
原文地址:https://www.cnblogs.com/FrankChen831X/p/11785429.html
时间: 2024-10-31 23:11:10