poj 1286 Necklace of Beads poj 2409 Let it Bead HDU 3923 Invoker <组合数学>

链接：http://poj.org/problem?id=1286

http://poj.org/problem?id=2409

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 typedef long long LL;
 LL P_M( LL a, LL b )
 {
     LL res=, t=a;
     while (b){
         if(b&)res*=t;
         t*=t;
         b>>=;
     }
     return res;
 }
 int a[], vi[];
 void dfs( int i )
 {
     if( !vi[i] ){
         vi[i]=;
         dfs(a[i]);
     }
 }

 int find(int t, int n ) // 求循环节 s == gcd( n, t ) ;
 {
     memset(vi, , sizeof vi);
     memset(a, , sizeof a);
     for( int i=; i<=n; ++ i ){
         int e=(i+t-)%n;
         if( e== )e=n;
         a[i]=e;
     }
     int s=;
    for( int i=; i<=n; ++ i ){
         if(!vi[i]){
             dfs(i);
             s++;
         }
     }
     return s;
 }
 int gcd( int x, int y )
 {
     return y==?x:gcd( y, x%y );
 }
 int main( )
 {
     int N;
     while( scanf("%d", &N)!= EOF){
         if( N==- )break;
         if( N== ){puts(""); continue;}
         LL ans=;
         if( !(N&) ){
             ans=P_M(3LL, N);
             for( int i=; i<=N; ++i ){
                 int t=find(i, N);
                 ans+=P_M( 3LL,t );
             }
             ans+=(LL)(N/)*(P_M(3LL,(N+)/));
             ans+=(LL)(N/)*P_M(3LL,N/);
         }
         else{
             for( int i=; i<N; ++i ){
                ans+=P_M(3LL, gcd(i, N));
             }
             ans+=(LL)(N)*(P_M(3LL,N/+));

         }
         printf( "%d\n", ans/(*N) );
     }
     return ;
 }

poj 1286

 #include <cstdio>
 #include <iostream>
 using namespace std;
 int N, M;
 typedef long long LL;
 LL P_M(int a, int b )
 {
     LL res=,t=a;
     while(b){
         if(b&)res*=t;
         t*=t;
         b>>=;
     }
     return res;
 }
 int gcd( int x, int y )
 {
     return y==?x:gcd( y, x%y );
 }
 LL polya( int k, int m )
 {
     LL ans=;
     for( int i=; i<m; ++ i ){
         ans+=P_M(k, gcd( i, m ));
     }
     if( m& ){
         ans+=(LL)(m*P_M( k, m/+ ) );
     }
     else {
         ans+=(LL)m/*P_M(k, m/);
         ans+=(LL)m/*P_M(k, m/+);
     }
     if(m)ans/=,ans/=m;
     return ans;
 }
 int main( )
 {
     while(scanf("%d%d", &N,&M)!=EOF, N+M ){
         printf("%lld\n", polya(N, M));
     }
     return ;
 }

poj 2409

 #include <cstdio>
 #include <iostream>
 using namespace std;
 typedef long long LL;
 const LL M=1e9+;
 LL P_M(int a, int b )
 {
     LL res=,t=a;
     while(b){
         if(b&)res*=t, res%=M;
         t*=t, t%=M;
         b>>=;
     }
     return res;
 }
 int gcd( int x, int y )
 {
     return y==?x:gcd( y, x%y );
 }
 LL polya( int k, int m )
 {
     LL ans=;
     for( int i=; i<m; ++ i ){
         ans+=P_M(k, gcd( i, m ));
         ans%=M;
     }
     if( m& ){
         ans+=(LL)(m*P_M( k, m/+ ) );
     }
     else {
         ans+=(LL)m/*P_M(k, m/);
         ans+=(LL)m/*P_M(k, m/+);
     }
     ans%=M;
     return   (ans*P_M(*m,M-))%M;// 求逆元
 }
 int main( )
 {
     int T, N, K, t=;
     scanf("%d", &T);
     while(T--){
         scanf("%d%d", &N,&K);
         printf("Case #%d: %I64d\n", t++, polya(N, K));
     }
     return ;
 }

hdu 3923