ZeptoLab Code Rush 2015 D. Om Nom and Necklace [题意] 给出一个字符串s,判断其各个前缀是否是 ABABA…ABA的形式(A和B都可以为空,且A有Q+1个,B有Q个,Q给定). [官方题解] 对于前缀P,我们可以把它拆成P=SSSS…SSSST,其中T是S的前缀.显然可以用KMP算法,时间复杂度是O(n). 当T=S:P=SSS…S.假设S出现了R次.如果转换为ABABAB…ABABA的形式,A和B是由几个S组成,而且最后的A一定是P的一个后缀.由…
D. Om Nom and Necklace time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output One day Om Nom found a thread with n beads of different colors. He decided to cut the first several beads from this th…
D. Om Nom and Necklace time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output One day Om Nom found a thread with n beads of different colors. He decided to cut the first several beads from this th…
poj2406 Power Strings(kmp) 给出一个字符串,问这个字符串是一个字符串重复几次.要求最大化重复次数. 若当前字符串为S,用kmp匹配'\0'+S和S即可. #include <cstdio> #include <cstring> using namespace std; const int maxn=2e6+5; char s1[maxn], s2[maxn]; int n1, n2, nxt[maxn], ans; int main(){ while (~…
C - Om Nom and Candies 思路:贪心+思维(或者叫数学).假设最大值max(wr,wb)为wr,当c/wr小于√c时,可以枚举r糖的数量(从0到c/wr),更新答案,复杂度√c:否则,假设hr/wr<hb/wr,得到hr*wb<hb*wr,由这个等式可知,在有wb*wr重量限制的情况下,买wb个r糖没有买wr个b糖划算,当需要买超过wb个r糖时,不如去买b糖,可以枚举r糖的数量(从0到wb-1),更新答案,复杂度√c. 代码: #include<bits/stdc++…
Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative inte…
