全部博文(68)
分类: C/C++
2012-02-02 21:25:35
Consider all integer combinations of ab for 2 a 5 and 2 b 5:
22=4, 23=8, 24=16, 25=32
32=9, 33=27, 34=81, 35=243
42=16, 43=64, 44=256, 45=1024
52=25, 53=125, 54=625, 55=3125
If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
How many distinct terms are in the sequence generated by ab for 2 a 100 and 2 b 100?
答案:9183
duplication
4 --> 49
8 --> 49
9 --> 49
16 --> 58
25 --> 49
27 --> 49
32 --> 48
36 --> 49
49 --> 49
64 --> 62
81 --> 58
100 --> 49 sum = 618
99 50 50 41 51 37
2 4 8 16 32 64 //Use the set data structure, count the number of 2's exponent.
3 9 27 81
5 25
6 36
7 49
10 100
result = 99*99-618 = 9183