FIB Query, 1007, 2011 Multi-University Training Contest 10

FIB Query, 1007, 2011 Multi-University Training Contest 10

FIB Query

TimeLimit: 2 Second   MemoryLimit: 64 Megabyte

Totalsubmit: 733   Accepted: 87  

Description

We all know the definition of Fibonacci series: fib[i]=fib[i-1]+fib[i-2],fib[1]=1,fib[2]=1.And we define another series P associated with the Fibonacci series: P[i]=fib[4*i-1].Now we will give several queries about P:give two integers L,R, and calculate ∑P[i](L <= i <= R).

Input

There is only one test case.
The first line contains single integer Q – the number of queries. (Q<=10^4)
Each line next will contain two integer L, R. (1<=L<=R<=10^12)

Output

For each query output one line.
Due to the final answer would be so large, please output the answer mod 1000000007.

Sample Input

2
1 300
2 400

Sample Output

838985007
352105429

Source

[p][/p]

p[1] + p[2] + ... + p[n] = f[1]^2 + f[2]^2 + ... + f[2*n-1]^2 + f[2*n]^2 = f[2*n] * f[2*n+1]

 1  #include  < iostream >
 2  #include  < cstring >
 3 
 4  using   namespace  std;
 5 
 6  #define   MOD  1000000007
 7 
 8  typedef   long   long   I64;
 9 
10  void  getF( I64 n, I64  * fn, I64  * fnp1 ) {
11          I64 m[  2  ][  2  ], t[  2  ][  2  ], p[  2  ][  2  ];
12           int  i, j, k;
13 
14          m[  0  ][  0  ]  =   1 ;
15          m[  0  ][  1  ]  =   0 ;
16          m[  1  ][  0  ]  =   0 ;
17          m[  1  ][  1  ]  =   1 ;
18 
19          p[  0  ][  0  ]  =   1 ;
20          p[  0  ][  1  ]  =   1 ;
21          p[  1  ][  0  ]  =   1 ;
22          p[  1  ][  1  ]  =   0 ;
23 
24           while  ( n  !=   0  ) {
25                   if  ( n  &   1  ) {
26                           for  ( i  =   0 ; i  <   2 ;  ++ i ) {
27                                   for  ( j  =   0 ; j  <   2 ;  ++ j ) {
28                                          t[ i ][ j ]  =   0 ;
29                                           for  ( k  =   0 ; k  <   2 ;  ++ k ) {
30                                                  t[i][j]  =  (t[i][j] + m[i][k] * p[k][j])  %  MOD;
31                                          }
32                                  }
33                          }
34                           for  ( i  =   0 ; i  <   2 ;  ++ i ) {
35                                   for  ( j  =   0 ; j  <   2 ;  ++ j ) {
36                                          m[ i ][ j ]  =  t[ i ][ j ];
37                                  }
38                          }
39                  }
40 
41                  n  >>=   1 ;
42                   for  ( i  =   0 ; i  <   2 ;  ++ i ) {
43                           for  ( j  =   0 ; j  <   2 ;  ++ j ) {
44                                  t[ i ][ j ]  =   0 ;
45                                   for  ( k  =   0 ; k  <   2 ;  ++ k ) {
46                                          t[i][j]  =  (t[i][j] + p[i][k] * p[k][j])  %  MOD;
47                                  }
48                          }
49                  }
50                   for  ( i  =   0 ; i  <   2 ;  ++ i ) {
51                           for  ( j  =   0 ; j  <   2 ;  ++ j ) {
52                                  p[ i ][ j ]  =  t[ i ][ j ];
53                          }
54                  }
55          }
56           * fnp1  =  m[  0  ][  0  ];
57           * fn    =  m[  1  ][  0  ];
58  }
59 
60  void  getP( I64 n, I64  * pn ) {
61          I64 fn, fnp1;
62          n  =  n  +  n;
63          getF( n,  & fn,  & fnp1 );
64           * pn  =  ( fn  *  fnp1 )  %  MOD;
65  }
66 
67  int  main() {
68           int  q;
69          I64 le, ri, pl, pr;
70          cin  >>  q;
71           while  ( q --   >   0  ) {
72                  cin  >>  le  >>  ri;
73                  getP( le - 1 ,  & pl );
74                  getP( ri,  & pr );
75                  cout  <<  ( ( pr  +  MOD  -  pl )  %  MOD )  <<   " \n " ;
76          }
77           return   0 ;
78  }
79