Difference between revisions of "ICPC Asia-Kharagpur Onsite Contest 2019"
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Solved by yanghong. 04:31 (+1) | Solved by yanghong. 04:31 (+1) | ||
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+ | 结论题, k>0 时删除两条边 | ||
== Problem D == | == Problem D == | ||
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Solved by yanghong. 01:00 (+3) | Solved by yanghong. 01:00 (+3) | ||
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+ | 随机算法。 每次询问三个数, 有四种情况: | ||
+ | |||
+ | 1. 0, 所以三个数全0, 则相当于只用一次询问就得到了三个数 | ||
+ | |||
+ | 2. 3, 三个数全1, 只用一次询问 | ||
+ | |||
+ | 3. 1,随机顺序询问三个数中的数, 期望2.5次得到三个数 | ||
+ | |||
+ | 4. 2, 期望2.5次得到三个数 | ||
+ | |||
+ | 总的期望询问次数少于90000次 | ||
== Problem G == | == Problem G == | ||
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Solved by yanghong. 01:13 (+1) | Solved by yanghong. 01:13 (+1) | ||
+ | |||
+ | 负超几何分布 | ||
== Problem L == | == Problem L == |
Latest revision as of 15:13, 19 October 2020
Problem A
Solved by Once. 00:21 (+)
可能碰撞的两个小球的速度和位置的乘积是相同的。用 set 维护之后枚举即可。
Problem B
Solved by bingoier. 01:49 (+)
打表找规律题,发现小的往大的中间交错插入是最优的。
Problem C
Solved by yanghong. 04:31 (+1)
结论题, k>0 时删除两条边
Problem D
Unsolved. (-3)
Problem E
Solved by bingoier. 03:56 (+1)
因为每种颜色之间相互独立,可以分开考虑。
维护每两个相同颜色之间的距离,一正一反做个FFT取大的一半即是对答案的贡献。
Problem F
Solved by yanghong. 01:00 (+3)
随机算法。 每次询问三个数, 有四种情况:
1. 0, 所以三个数全0, 则相当于只用一次询问就得到了三个数
2. 3, 三个数全1, 只用一次询问
3. 1,随机顺序询问三个数中的数, 期望2.5次得到三个数
4. 2, 期望2.5次得到三个数
总的期望询问次数少于90000次
Problem G
Solved by Once. 00:30 (+)
枚举坐不坐火车的情况,计算。
Problem H
Unsolved.
Problem I
Unsolved.
Problem J
Solved by bingoier. 00:26 (+)
二分签到题。
Problem K
Solved by yanghong. 01:13 (+1)
负超几何分布
Problem L
Solved by Once. 02:26 (+1)
二分答案,之后由于体积只有 2 和 1 的情况,所以可以贪心,先放体积为 2 的,再放体积为 1 的。