Here's the problem from last year's ACM ICPC Dhaka Regional Preliminary Round.
It says to calculate sod(n) [sum of digits of n] until n >= 10. Here n = a^b, (0<= a,b <= 10^50,000 and a+b>0 ) Any thoughts on how to solve this?
# | User | Rating |
---|---|---|
1 | tourist | 3845 |
2 | jiangly | 3707 |
3 | Benq | 3630 |
4 | orzdevinwang | 3573 |
5 | Geothermal | 3569 |
5 | cnnfls_csy | 3569 |
7 | jqdai0815 | 3532 |
8 | ecnerwala | 3501 |
9 | gyh20 | 3447 |
10 | Rebelz | 3409 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 171 |
2 | adamant | 163 |
3 | awoo | 161 |
4 | maroonrk | 152 |
4 | nor | 152 |
6 | -is-this-fft- | 151 |
7 | TheScrasse | 148 |
8 | atcoder_official | 146 |
9 | Petr | 145 |
10 | pajenegod | 144 |
Repeated Digit Sum of A^B
Here's the problem from last year's ACM ICPC Dhaka Regional Preliminary Round.
It says to calculate sod(n) [sum of digits of n] until n >= 10. Here n = a^b, (0<= a,b <= 10^50,000 and a+b>0 ) Any thoughts on how to solve this?
Rev. | Lang. | By | When | Δ | Comment | |
---|---|---|---|---|---|---|
en2 |
![]() |
simplecomplex | 2018-06-06 14:09:23 | 6 | Tiny change: 's of n]** until **n >= 10' -> 's of n]** while **n >= 10' | |
en1 |
![]() |
simplecomplex | 2018-06-06 14:07:06 | 340 | Initial revision (published) |
Name |
---|