fibonacci数的递归求法：
1 2 struct Fib 3 { 4 int x1; 5 int x2; 6 }; 7 8 Fib fib(int x) 9 { 10 Fib ans; 11 if(x == 2) 12 { 13 ans.x1 = 0; 14 ans.x2 = 1%MOD; 15 return ans; 16 } 17 Fib temp; 18 int x1 = x/2, x2 = x1 - 1; 19 if(x1%2) 20 { 21 temp = fib(x2); 22 ans.x1 = (2*temp.x1 + temp.x2)*temp.x2%MOD; 23 ans.x2 = ((temp.x1 + temp.x2)*(temp.x1 + temp.x2) + temp.x2*temp.x2)%MOD; 24 } 25 else 26 { 27 temp = fib(x1); 28 ans.x1 = (temp.x1*temp.x1 + temp.x2*temp.x2)%MOD; 29 ans.x2 = (2*temp.x1 + temp.x2)*temp.x2%MOD; 30 } 31 return ans; 32 }