Given a range [l r] find the sum of last digits of fibonacci numbers between l and r. i.e , last(fib(l))+last(fib(l+1)) .... last(fib(r)),
where last(fib(a)) denotes last digit of ath fibonacci number.
1<=l<=r<=1e9
# | User | Rating |
---|---|---|
1 | jiangly | 3640 |
2 | Benq | 3593 |
3 | tourist | 3572 |
4 | orzdevinwang | 3561 |
5 | cnnfls_csy | 3539 |
6 | ecnerwala | 3534 |
7 | Radewoosh | 3532 |
8 | gyh20 | 3447 |
9 | Rebelz | 3409 |
10 | Geothermal | 3408 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 174 |
2 | awoo | 164 |
3 | adamant | 163 |
4 | TheScrasse | 159 |
5 | nor | 158 |
6 | maroonrk | 156 |
7 | -is-this-fft- | 151 |
8 | SecondThread | 147 |
9 | orz | 146 |
10 | pajenegod | 145 |
Given a range [l r] find the sum of last digits of fibonacci numbers between l and r. i.e , last(fib(l))+last(fib(l+1)) .... last(fib(r)),
where last(fib(a)) denotes last digit of ath fibonacci number.
1<=l<=r<=1e9
Name |
---|
Using Pisano Period for $$$base = 10$$$ then we have $$$\pi(n) = 60$$$ numbers per cycle