标签:think res idt math air rom problem rip font
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 39658 | Accepted: 16530 |
Description
Input
Output
Sample Input
abcd aaaa ababab .
Sample Output
1 4 3
Hint
Source
KMP算法
设字符串的长度为l,假设l%(l-next[l])=0。该序列为循环序列,循环节长度为l-next[l],答案即为l/(l-next[l]);反之则不为循环序列。答案为1。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define F(i,j,n) for(int i=j;i<=n;i++)
#define D(i,j,n) for(int i=j;i>=n;i--)
#define ll long long
#define pa pair<int,int>
#define maxn 1000010
#define inf 1000000000
using namespace std;
char s[maxn];
int l,nxt[maxn];
inline void getnext()
{
int i=0,j=-1;
nxt[0]=-1;
while (i<l)
{
if (j==-1||s[i]==s[j]) nxt[++i]=++j;
else j=nxt[j];
}
}
int main()
{
scanf("%s",s);
while (s[0]!='.')
{
l=strlen(s);
getnext();
if (l%(l-nxt[l])==0) printf("%d\n",l/(l-nxt[l]));
else printf("1\n");
scanf("%s",s);
}
}
标签:think res idt math air rom problem rip font
原文地址:http://www.cnblogs.com/claireyuancy/p/7079795.html