poj 2406 Power Strings(kmp)
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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
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.
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
题意:给你定义字符串的幂 然后给你一个串s 问 幂次最大为多少
思路:找到字符串的循环节 然后看下有没有满足的循环节 没有就输出1
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<string> #include<vector> #include<stack> #include<bitset> #include<cstdlib> #include<cmath> #include<set> #include<list> #include<deque> #include<map> #include<queue> #define ll long long int using namespace std; inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;} int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1}; int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1}; const int inf=0x3f3f3f3f; const ll mod=1e9+7; char s[1000007]; int nextt[1000007]; void getnext(){ nextt[1]=0; int len=strlen(s); for(int i=2,j=0;i<=len;i++){ while(j>0&&s[i-1]!=s[j]) j=nextt[j]; if(s[i-1]==s[j]) j++; nextt[i]=j; } } int main(){ //ios::sync_with_stdio(false); while(~scanf("%s",s)){ if(strlen(s)==1&&s[0]==‘.‘) break; getnext(); int len=strlen(s); int temp=nextt[len]; while(len%(len-temp)!=0){ temp=nextt[temp]; if(len/(len-temp)<2) break; } if(len%(len-temp)==0) printf("%d ",len/(len-temp)); else printf("1 "); } return 0; }
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