solution is that check if there is an entry in the array from i=2 to i=n-1; such that a[i]<a[i-1]&&a[i]<a[i+1]. If there is such a value ans is NO else answer is always "YES". This is an easy way to solve that problem.... OK................
# | User | Rating |
---|---|---|
1 | jiangly | 3976 |
2 | tourist | 3815 |
3 | jqdai0815 | 3682 |
4 | ksun48 | 3614 |
5 | orzdevinwang | 3526 |
6 | ecnerwala | 3514 |
7 | Benq | 3482 |
8 | hos.lyric | 3382 |
9 | gamegame | 3374 |
10 | heuristica | 3357 |
# | User | Contrib. |
---|---|---|
1 | cry | 169 |
2 | -is-this-fft- | 165 |
3 | Um_nik | 161 |
3 | atcoder_official | 161 |
5 | djm03178 | 157 |
6 | Dominater069 | 156 |
7 | adamant | 154 |
8 | luogu_official | 152 |
9 | awoo | 151 |
10 | TheScrasse | 147 |
hint to solve pillars problem educational codeforces round division 2.
solution is that check if there is an entry in the array from i=2 to i=n-1; such that a[i]<a[i-1]&&a[i]<a[i+1]. If there is such a value ans is NO else answer is always "YES". This is an easy way to solve that problem.... OK................
Rev. | Lang. | By | When | Δ | Comment | |
---|---|---|---|---|---|---|
en2 | AM_I_Learning | 2019-07-23 13:50:22 | 23 | Tiny change: 'at problem' -> 'at problem.... OK................' (published) | ||
en1 | AM_I_Learning | 2019-07-23 13:49:36 | 291 | Initial revision (saved to drafts) |
Name |
---|