HDU 5171 GTY's birthday gift（矩阵快速幂 ）

 

 

HDU 5171 GTY's birthday gift ( 矩阵快速幂裸题目 )

隐藏行号 复制代码 ？ 矩阵快速幂

1. #include <cstdio>    
   
   

2. #include <cstring>
   
   

3. #include <algorithm>
   
   

4. using namespace std;
   
   

5. #define MAX_SIZE 3
   
   

6. #define MOD 10000007
   
   

7. #define clr( a, b ) memset( a, b, sizeof(a) )
   
   

8. typedef long long LL;
   
   

10. struct Mat
    
    

11. {
    
    

12.     int n;
    
    

13.     LL mat[MAX_SIZE][MAX_SIZE];
    
    

14.     Mat( int _n )
    
    

15.     {
    
    

16.         n = _n;
    
    

17.         clr( mat, 0 );
    
    

18.     }
    
    

19.     void init()
    
    

20.     {
    
    

21.         for( int i = 0; i < n; ++i )
    
    

22.             for( int j = 0; j < n; ++j )
    
    

23.                 mat[i][j] = ( i == j );
    
    

24.     }
    
    

25.     Mat operator * ( const Mat &b ) const
    
    

26.     {
    
    

27.         Mat c( b.n );
    
    

28.         for( int k = 0; k < n; ++k )
    
    

29.             for( int i = 0; i < n; ++i )    if( mat[i][k] )
    
    

30.                 for( int j = 0; j < n; ++j )
    
    

31.                     c.mat[i][j] = ( c.mat[i][j] + mat[i][k] * b.mat[k][j] ) % MOD;
    
    

32.         return c;
    
    

33.     }
    
    

34. };
    
    

36. Mat fast_mod( Mat a, int b )
    
    

37. {
    
    

38.     Mat res( a.n );
    
    

39.     res.init();
    
    

40.     while( b )
    
    

41.     {
    
    

42.         if( b & 1 )    res = res * a;
    
    

43.         a = a * a;
    
    

44.         b >>= 1; 
    
    

45.     }
    
    

46.     return res;
    
    

47. }
    
    

49. void scan( int &x )
    
    

50. {
    
    

51.     char c;
    
    

52.     while( c = getchar(), c < '0' || c > '9' );
    
    

53.     x = c - '0';
    
    

54.     while( c = getchar(), c >= '0' && c <= '9' ) x = x *10 + c - '0';
    
    

55. }
    
    

57. int n, k, x[ 100001 ], id;
    
    

58. int first, second;
    
    

59. LL sum, c[ 3 ], res;
    
    

61. int main()
    
    

62. {
    
    

63.     Mat M( 3 );
    
    

64.     M.mat[0][0] = M.mat[0][2] = M.mat[1][2] = M.mat[2][1] = M.mat[2][2] = 1;
    
    

65.     while( ~scanf( "%d %d", &n, &k ) )
    
    

66.     {
    
    

67.         sum = 0;
    
    

68.         first = second = -1;
    
    

69.         for( int i = 0; i < n; ++i )
    
    

70.         {
    
    

71.             scan( x[i] );
    
    

72.             sum += x[i];
    
    

73.             if( sum > MOD ) sum -= MOD; 
    
    

74.             if( x[i] > first )
    
    

75.             {
    
    

76.                 first = x[i];
    
    

77.                 id = i;
    
    

78.             }
    
    

79.         }
    
    

80.         for( int i = 0; i < n; ++i )
    
    

81.         {
    
    

82.             if( i == id ) continue;
    
    

83.             if( x[i] > second ) second = x[i];
    
    

84.         }
    
    

85.         c[ 0 ] = sum - first;
    
    

86.         c[ 1 ] = second;
    
    

87.         c[ 2 ] = first;
    
    

88.     //    printf( "c[0] = %lld c[1] =  %lld c[2] = %lld\n", c[0],c[1],c[2] );
    
    

89.         Mat b = fast_mod( M, k + 1 );
    
    

90.         res = ( c[ 0 ] *  b.mat[0][0] + c[ 1 ] * b.mat[0][1] + c[ 2 ] * b.mat[0][2] ) % MOD ;
    
    

91.         printf( "%I64d\n", res );
    
    

93.     }
    
    

94.     return 0;
    
    

95. }