poj 2409 Let it Bead (polya)
2015-04-12 10:37
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#include<iostream> #include<math.h> #include<stdio.h> #include<algorithm> #include<string.h> #include<vector> #include<queue> #include<map> #include<set> #define B(x) (1<<(x)) using namespace std; typedef long long ll; void cmax(int& a,int b){ if(b>a)a=b; } void cmin(int& a,int b){ if(b<a)a=b; } void cmax(ll& a,ll b){ if(b>a)a=b; } void cmin(ll& a,ll b){ if(b<a)a=b; } void add(int& a,int b,int mod){ a=(a+b)%mod; } void add(ll& a,ll b,ll mod){ a=(a+b)%mod; } const int oo=0x3f3f3f3f; const ll MOD=1000000007; ll quick_pow(ll a,ll k){ ll ans=1; while(k){ if(k&1) ans=(ans*a); a=(a*a); k>>=1; } return ans; } int gcd(int a,int b){ return b ? gcd(b,a%b) : a; } int main() { int c,n; while(scanf("%d %d",&c,&n)!=EOF){ if(n==0&&c==0)break; if(n==0){ printf("0\n"); continue; } ll ans=0; if(n&1){ ans=quick_pow(c,(n-1)/2+1)*n; for(int i=1;i<=n;i++) ans+=quick_pow(c,gcd(n,i)); }else{ ans=quick_pow(c,(n-2)/2+2)*n/2; ans+=quick_pow(c,n/2)*n/2; for(int i=1;i<=n;i++) ans+=quick_pow(c,gcd(n,i)); } cout<<ans/n/2<<endl; } return 0; }
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