As the title, 1e18 + 1 == 1e18 in C++.
This is because the double type stores the first 53 binary bits of a number. So the number 1 will be ignored.
And most of submissions which fst on test 13 of problem C of edu round 187 are for this reason.
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1e18 + 1 == 1e18 in C++
As the title, 1e18 + 1 == 1e18 in C++.
This is because the double type stores the first 53 binary bits of a number. So the number 1 will be ignored.
And most of submissions which fst on test 13 of problem C of edu round 187 are for this reason.
| Rev. | Язык | Кто | Когда | Δ | Комментарий | |
|---|---|---|---|---|---|---|
| en4 |
|
XiangXunyi | 2026-03-07 15:50:14 | 38 | Add a suggestion that using `1'000'000'000'000'000'001LL` | |
| en3 |
|
XiangXunyi | 2026-02-27 16:24:33 | 193 | ||
| en2 |
|
XiangXunyi | 2026-02-26 10:43:28 | 7 | Tiny change: 'irst 53 bits of a' -> 'irst 53 binary bits of a' | |
| en1 |
|
XiangXunyi | 2026-02-26 10:41:21 | 268 | Initial revision (published) |
| Название |
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