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

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 using namespace std;
 7 typedef long long LL;
 8 LL P_M( LL a, LL b )
 9 {
10     LL res=1, t=a;
11     while (b){
12         if(b&1)res*=t;
13         t*=t;
14         b>>=1;
15     }
16     return res;
17 }
18 int a[30], vi[30];
19 void dfs( int i )
20 {
21     if( !vi[i] ){
22         vi[i]=1;
23         dfs(a[i]);
24     }
25 }
26 
27 int find(int t, int n ) // 求循环节 s == gcd( n, t ) ;
28 {
29     memset(vi, 0, sizeof vi);
30     memset(a, 0, sizeof a);
31     for( int i=1; i<=n; ++ i ){
32         int e=(i+t-1)%n;
33         if( e==0 )e=n;
34         a[i]=e;
35     }
36     int s=0;
37    for( int i=1; i<=n; ++ i ){
38         if(!vi[i]){
39             dfs(i);
40             s++;
41         }
42     }
43     return s;
44 }
45 int gcd( int x, int y )
46 {
47     return y==0?x:gcd( y, x%y );
48 }
49 int main( )
50 {
51     int N;
52     while( scanf("%d", &N)!= EOF){
53         if( N==-1 )break;
54         if( N==0 ){puts("0"); continue;}
55         LL ans=0;
56         if( !(N&1) ){
57             ans=P_M(3LL, N);
58             for( int i=2; i<=N; ++i ){
59                 int t=find(i, N);
60                 ans+=P_M( 3LL,t );
61             }
62             ans+=(LL)(N/2)*(P_M(3LL,(N+2)/2));
63             ans+=(LL)(N/2)*P_M(3LL,N/2);
64         }
65         else{
66             for( int i=0; i<N; ++i ){
67                ans+=P_M(3LL, gcd(i, N));
68             }
69             ans+=(LL)(N)*(P_M(3LL,N/2+1));
70 
71         }
72         printf( "%d
", ans/(2*N) );
73     }
74     return 0;
75 }

poj 1286

 1 #include <cstdio>
 2 #include <iostream>
 3 using namespace std;
 4 int N, M;
 5 typedef long long LL;
 6 LL P_M(int a, int b )
 7 {
 8     LL res=1,t=a;
 9     while(b){
10         if(b&1)res*=t;
11         t*=t;
12         b>>=1;
13     }
14     return res;
15 }
16 int gcd( int x, int y )
17 {
18     return y==0?x:gcd( y, x%y );
19 }
20 LL polya( int k, int m )
21 {
22     LL ans=0;
23     for( int i=0; i<m; ++ i ){
24         ans+=P_M(k, gcd( i, m ));
25     }
26     if( m&1 ){
27         ans+=(LL)(m*P_M( k, m/2+1 ) );
28     }
29     else {
30         ans+=(LL)m/2*P_M(k, m/2);
31         ans+=(LL)m/2*P_M(k, m/2+1);
32     }
33     if(m)ans/=2,ans/=m;
34     return ans;
35 }
36 int main( )
37 {
38     while(scanf("%d%d", &N,&M)!=EOF, N+M ){
39         printf("%lld
", polya(N, M));
40     }
41     return 0;
42 }

poj 2409

 1 #include <cstdio>
 2 #include <iostream>
 3 using namespace std;
 4 typedef long long LL;
 5 const LL M=1e9+7;
 6 LL P_M(int a, int b )
 7 {
 8     LL res=1,t=a;
 9     while(b){
10         if(b&1)res*=t, res%=M;
11         t*=t, t%=M;
12         b>>=1;
13     }
14     return res;
15 }
16 int gcd( int x, int y )
17 {
18     return y==0?x:gcd( y, x%y );
19 }
20 LL polya( int k, int m )
21 {
22     LL ans=0;
23     for( int i=0; i<m; ++ i ){
24         ans+=P_M(k, gcd( i, m ));
25         ans%=M;
26     }
27     if( m&1 ){
28         ans+=(LL)(m*P_M( k, m/2+1 ) );
29     }
30     else {
31         ans+=(LL)m/2*P_M(k, m/2);
32         ans+=(LL)m/2*P_M(k, m/2+1);
33     }
34     ans%=M;
35     return   (ans*P_M(2*m,M-2))%M;// 求逆元
36 }
37 int main( )
38 {
39     int T, N, K, t=1;
40     scanf("%d", &T);
41     while(T--){
42         scanf("%d%d", &N,&K);
43         printf("Case #%d: %I64d
", t++, polya(N, K));
44     }
45     return 0;
46 }

hdu 3923

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原文地址：https://www.cnblogs.com/jian1573/p/3233414.html