poj2406 Power Strings
阿新 • • 發佈:2017-06-26
lib ring concat accept std ont mod long long abcde
Power Strings
Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 39658 | Accepted: 16530 |
Description
Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).Input
Output
For each s you should print the largest n such that s = a^n for some string a.Sample Input
abcd aaaa ababab .
Sample Output
1 4 3
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.Source
Waterloo local 2002.07.01
KMP算法
設字符串的長度為l,假設l%(l-next[l])=0。該序列為循環序列,循環節長度為l-next[l],答案即為l/(l-next[l]);反之則不為循環序列。答案為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); } }
poj2406 Power Strings