Poj 2406 power strings (KMP)

Language:DefaultPower strings Time limit:3000 Ms   Memory limit:65536 K Total submissions:33335   Accepted:13852 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 shoshould print the largest N such that S = a ^ N for some string. Sample Input abcdaaaaababab. Sample output 143 Hint This problem has huge input, use scanf instead of CIN to avoid time limit exceed. Source Waterloo local 2002.07.01

Returns the minimum cycle of a string.

For a string, if abcdabc is next [Len] = 3, then Len-next [Len] is greater than Len/2, then Len % (LEN-next [Len])! = 0; for a periodic string abababab next [Len] = 4, len-next [Len] should be equal

The minimum length of the string, so if there is a minimum cycle, you can use Len % (LEN-next [Len]) to determine whether it is 0 (in my opinion, if there is a mistake, please tell me)

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8using namespace std;#define N 100005char a[N];int len,next[N];void getfail(char *a){ int i,j; len=strlen(a); i=0;j=-1; next[0]=-1; while(i<len) { if(j==-1||a[i]==a[j]){ i++; j++; next[i]=j;}elsej=next[j]; }}int main(){int i,j;while(scanf("%s",a)){if(a[0]=='.') break;getfail(a);int ans=len%(len-next[len]);if(ans==0)printf("%d\n",len/(len-next[len]));elseprintf("1\n");}return 0;}

Poj 2406 power strings (KMP)