标签：i++ ems problem test ace sam power ges last
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
Sample Input
abcd aaaa ababab .
Sample Output
1 4 3
Hint
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cstring> 5 6 using namespace std; 7 8 char s[1000005]; 9 int Next[1000005]; 10 11 void getnext(char* s,int m) 12 { 13 Next[0]=0; 14 Next[1]=0; 15 for(int i=1;i<m;i++) 16 { 17 int j=Next[i]; 18 while(j&&s[i]!=s[j]) 19 j=Next[j]; 20 if(s[i]==s[j]) 21 Next[i+1]=j+1; 22 else 23 Next[i+1]=0; 24 } 25 } 26 27 int main() 28 { 29 while(scanf("%s",s)) 30 { 31 if(s[0]==‘.‘&&s[1]==‘\0‘) 32 break; 33 memset(Next,0,sizeof(Next)); 34 int m=strlen(s); 35 getnext(s,m); 36 int x=1; 37 if(m%(m-Next[m])==0) 38 x=m/(m-Next[m]); 39 printf("%d\n",x); 40 } 41 42 43 return 0; 44 }
标签：i++ ems problem test ace sam power ges last
原文地址：http://www.cnblogs.com/xibeiw/p/7308256.html
