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| WA7 | Mortus | 1608. Счастливые билеты 2008 | 12 янв 2024 00:41 | 1 |
WA7 Mortus 12 янв 2024 00:41 |
| WA19 | Alexs | 1608. Счастливые билеты 2008 | 5 мар 2021 00:08 | 1 |
WA19 Alexs 5 мар 2021 00:08 .I'm already tired of solving this number..Mods please give me 19th test case... |
| Is 1001 unlucky or 1010? | roman velichkin | 1608. Счастливые билеты 2008 | 8 ноя 2019 10:25 | 1 |
I didn't get this task.
Is 1001 unlucky number or 1010? |
| WA 7, WA 8 | 💻Evgeny Nemtsev [UrFU FT-17]'` | 1608. Счастливые билеты 2008 | 10 авг 2017 22:53 | 1 |
WA 7, WA 8 💻Evgeny Nemtsev [UrFU FT-17]'` 10 авг 2017 22:53 |
| Give me some tests please WA 16 | Aneto | 1608. Счастливые билеты 2008 | 17 ноя 2015 19:08 | 4 |
There was an error with data input using StreamTokenizer, so be careful if you are using it.
Edited by author 05.04.2008 21:15
Edited by author 06.04.2008 04:02 Thank you! Scanner() got AC. int main() { int n; cin>>n; int i; for(i=0;i<n/2;i++) cout<<i+1<<" "<<n-i<<" "; if(n%2!=0) cout<<((int)n/2)+1<<" "; cout<<endl; cout<<"1 "; for(i=3;i<n;i+=2) cout<<i<<" "; int p=n; if(p%2!=0) { p-=1; cout<<n<<" "; } for(p=p;p>1;p-=2) cout<<p<<" "; return 0; } input1:
1651 1654654564
output1:
1043406233
intput2:
77251 78156464666454
output2:
37919407177871
intput3:
1 999999999999
output3:
537412247190
input4:
101231 1000000000000000000
output4:
391773528202165271
input5:
123456789 987654321
output5:
567000597
input6:
202250020115 3000200002203254
output6:
1339741370499750
|
| WA18 | Nisarg Shah | 1608. Счастливые билеты 2008 | 15 дек 2008 10:49 | 1 |
WA18 Nisarg Shah 15 дек 2008 10:49 I had WA7....corrected one error but now getting WA18...Mods please give me 18th test case...
Thanks...
Edited by author 09.01.2009 13:12 |
| How to solve this? I get TLE10 (-) | rohit | 1608. Счастливые билеты 2008 | 10 авг 2008 00:08 | 2 |
Count F(n) - number of lucky tickets in range 1..n, the answer is F(b) - F(a-1). If n=4567, then F(4567) is number of 1-digit lucky numbers + number of 2-digit lucky numbers + number of 3-digit lucky numbers. With 4 digit numbers it's a bit harder - if first digit is 1..3, then remainder can be anything sustaining luckiness (i.e. there is no effect from upper border). Also first digit can't be 5,6,7. If first digit is 4, the last one can be anything except 4 - so you have multiplier 9 for the next step. This is not everything, but the basic idea. |
| wa7 | zhang_xiao | 1608. Счастливые билеты 2008 | 10 авг 2008 00:04 | 4 |
wa7 zhang_xiao 31 мар 2008 15:33 boss
I wa7..........
I want data
zhangxiao19901124@gmail.com Re: wa7 zhang_xiao 31 мар 2008 18:35 Now Wa 10......
Can somebody E-mail me an AC program,I want to Find out the problem! Thanks a lot
zhangxiao19901124@gmail.com You can try use int64.
I use extended at first,but Wrong answer.
And use int64,I get AC. Re: wa7 Denis Koshman 10 авг 2008 00:04 Just compare it with brute force solution for first 10'000'000 values. Also check that values 10, 100, 1000, 10'000, ..., 10^18 come in ascdending order - this way you'll check for overflow.
Signed int64 is enough here if you perform no redundant additions/multiplications (i.e. if resulting values only increase during execution). |
