G - Power Strings
Time Limit:3000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 2406
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
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
1.直接暴力枚举
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#include
#include
#include
#include
#include
#include
using namespace std; const int maxn=1000005; typedef long long LL; char s[maxn]; int main(){ int T; //freopen("Text//in.txt","r",stdin); while(~scanf("%s",s)&&s[0]!='.'){ int len=strlen(s); for(int i=1;i<=len;i++)if(len%i==0){ int ok=1; for(int j=i;j
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2.kmp算法 i-next[i]:表示已i结尾的前缀的循环节的长度
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#include
#include
#include
#include
#include
#include
using namespace std; const int maxn=1000005; typedef long long LL; char s[maxn]; int next[maxn]; void getnext(char*s){ int len=strlen(s); int i=0,j=-1; next[0]=-1; while(i
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