Common Divisors CodeForces, divisorscodeforces

Common Divisors CodeForces, divisorscodeforces

Common Divisors CodeForces-182D

Idea: Use kmp to find the next array to find the minimum cycle length (http://blog.csdn.net/acraz/article/details/47663477,http://www.cnblogs.com/chenxiwenruo/p/3546457.html) of two strings, and then get the minimum cycle, if the minimum cycle is different, the answer is 0, otherwise, the minimum number of cycle segments contained in each string is obtained, and the number of public factors (that is, the number of factors of the maximum public factor) of these two numbers is the answer.

 1 #include<cstdio> 2 #include<cmath> 3 #include<cstring> 4 char s1[100010],s2[100010]; 5 char c[100010],d[100010]; 6 int f[100010]; 7 int m1,m2,ans; 8 int getf(char *P,int *f,int& m) 9 {10     int j=f[0]=-1,i=0;11     while(i<m)12     {13         while(j>=0&&P[i]!=P[j])    j=f[j];14         ++j;++i;15         f[i]=j;16     }17     return m%(m-f[m])==0?m-f[m]:m;18 }19 int gcd(int a,int b)20 {21     int t;22     while(b!=0)23     {24         t=a;25         a=b;26         b=t%b;27     }28     return a;29 }30 int main()31 {32     int i,p,t;33     scanf("%s%s",s1,s2);34     m1=strlen(s1);35     m2=strlen(s2);36     int a=getf(s1,f,m1);37     int b=getf(s2,f,m2);38     strncpy(c,s1,a);39     strncpy(d,s2,b);40     if(strcmp(c,d)!=0)41         printf("0");42     else43     {44         p=gcd(m1/a,m2/b);45         t=(int)sqrt(p+0.5);46         for(i=1;i<=t;i++)47             if(p%i==0)48                 ans+=2;49         if(t*t==p)    ans--;50         printf("%d",ans);51     }52     return 0;53 }