POJ 2406 Power Strings(KMP+最小循环节)
2017-08-21 15:14
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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
Sample Output
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
题解:
给一个串,问是由最多几个循环节组成的,直接用最小循环节长度len-next[len]判断一下就好了,如果能整除len就输出len/最小循环节长度,否则输出1
代码:
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.
题解:
给一个串,问是由最多几个循环节组成的,直接用最小循环节长度len-next[len]判断一下就好了,如果能整除len就输出len/最小循环节长度,否则输出1
代码:
#include<algorithm> #include<iostream> #include<cstring> #include<stdio.h> #include<math.h> #include<string> #include<stdio.h> #include<queue> #include<stack> #include<map> #include<vector> #include<deque> using namespace std; #define lson k*2 #define rson k*2+1 #define M (t[k].l+t[k].r)/2 #define INF 1008611111 #define ll long long #define eps 1e-15 int Next[1000005]; char T[1000005]; void init() { int i=0,j=-1; Next[0]=-1; while(T[i]) { if(j==-1||T[i]==T[j]) { i++; j++; Next[i]=j; } else j=Next[j]; } } int main() { int i,j,n,len; while(gets(T)!=NULL) { if(T[0]=='.') break; init(); len=strlen(T); n=len-Next[len]; if(len%n==0) printf("%d\n",len/n); else printf("1\n"); } return 0; }
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