Difference between revisions of "2018 CCPC Online Contest"
Line 37: | Line 37: | ||
Solved by oxx1108. | Solved by oxx1108. | ||
− | 题意:求$(A+\sqrt{B})^n$ | + | 题意:求$(A+\sqrt{B})^n$二项式定理展开的奇数项的和(含根号项)$mod p$ |
− | + | 题解:答案为$((A+\sqrt{B})^n - (A+\sqrt{B})^n) / 2$,实现一下带根号的乘法然后快速幂即可。 | |
+ | 注意不能直接求2的逆元,因为题目没有保证时素数,所以对$p * 2$取模最后再对答案除2即可。 | ||
== Problem F == | == Problem F == |
Revision as of 13:45, 25 August 2018
oxx
Problem A
Solved by dreamcloud.
题意:
题解:
Problem B
Unsolved.
题意:
题解:
Problem C
Solved by oxx1108.
题意:在环中重载加法和乘法,使得在满足$(m + n) ^ p = m ^ p + n ^ p$ $p$是素数
题解:直接重载取模意义下的加和乘就行了。证明费马小定理。
Problem D
Solved by oxx1108.
题意:给定$a$和$n$求$b$和$c$,求满足$a ^ n + b ^ n = c ^ n$ 一组$b$,$c$
题解:n一定等于1或者2,然后1的时候显然,2的时候勾股数构造一下就可以了。($a$为2时平方无解,但是出题人友好地去掉了这个情况)
Problem E
Solved by oxx1108.
题意:求$(A+\sqrt{B})^n$二项式定理展开的奇数项的和(含根号项)$mod p$
题解:答案为$((A+\sqrt{B})^n - (A+\sqrt{B})^n) / 2$,实现一下带根号的乘法然后快速幂即可。 注意不能直接求2的逆元,因为题目没有保证时素数,所以对$p * 2$取模最后再对答案除2即可。
Problem F
Unsolved.
题意:
题解:
Problem G
Unsolved.
题意:
题解:
Problem H
Unsolved.
题意:
题解:
Problem I
Solved by oxx1108.
题意:
题解:
Problem J
Solved by dreamcloud.
题意:
题解:
ECNU Foreigners
Problem A
Solved by kblack. 00:49 (+)
Problem B
Unsolved.
Problem C
Solved by zerol. 00:50 (+)
Problem D
Solved by zerol. 00:29 (+)
Problem E
Solved by ultmaster. 02:50 (+1)
题意:求 $\displaystyle \sum_{i=1}^n [i \equiv 1 \pmod 2] \frac{a^{n-i} \sqrt{b}^i n!}{i! (n-i)!}$。
题解:简单观察,或者使用 Mathematica 化简之后就可以得到所求式是 $\displaystyle \frac{1}{2} \left( (a + \sqrt{b})^n - (a - \sqrt{b})^n \right)$。发现 $(a+\sqrt{b})^n$ 一定具有 $a + c \sqrt{b}$ 的形式,直接快速幂即可。最后要把 $\sqrt{b}$ 中的平方因子提出来,暴力搞好像会 TLE(可能是前面模多了),要预处理。
处理模数要用到经典的技巧:先模 $2p$,然后最后除以 2。
Problem F
Unsolved. (-1)
Problem G
Solved by zerol. 03:03 (+3)
Problem H
Unsolved.
Problem I
Solved by kblack. 00:26 (+1)
Problem J
Solved by ultmaster. 00:36 (+)
题意:有若干格点,格点上有积分,积分只有从 $(x-1,y-1)$ 走过去才能得到,求最高积分。
题解:把它想成一列一列的格子一起转移;按 $x$ 升序 $y$ 降序排,然后由于 $y$ 降序了,所以更新上面的东西并不会影响后面的。直接用一个线段树支持 set 标记和查询区间最大即可。