32x |
3+32x |
32−2x |
3+32−2x |
32x |
3+32x |
32−2x•32x−1 |
(3+32−2x)•32x−1 |
32x |
3+32x |
3 |
3+32x |
故f(
1 |
101 |
100 |
101 |
2 |
101 |
99 |
101 |
故f(
1 |
101 |
2 |
101 |
100 |
101 |
故答案为:50
32x |
3+32x |
1 |
101 |
2 |
101 |
100 |
101 |
32x |
3+32x |
32−2x |
3+32−2x |
32x |
3+32x |
32−2x•32x−1 |
(3+32−2x)•32x−1 |
32x |
3+32x |
3 |
3+32x |
1 |
101 |
100 |
101 |
2 |
101 |
99 |
101 |
1 |
101 |
2 |
101 |
100 |
101 |