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

 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\n", 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\n", 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\n", t++, polya(N, K));

44     }

45     return 0;

46 }

hdu 3923