数学技巧真有趣,看出规律就很简单了 wa 题意:给出数k  输出所有阶乘尾数有k个0的数 这题来来回回看了两三遍, 想的方法总觉得会T 后来想想  阶乘 emmm  1*2*3*4*5*6*7*8*9*10...*n 尾数的0只与5有关 是5的几倍就有几个0  因为5前面肯定有偶数 乘起来就有一个0 而且最后输出肯定是连续的5个 hhh 兴奋 开始上手 乱搞一下  发现复杂度还行 测样例 发现 k=5 的时候不对了 输出25~29了 应该是0的 咦  测了一下 25!应该是6个0的 25=5*5…
B. A Trivial Problem 题目连接: http://www.codeforces.com/contest/633/problem/B Description Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such t…
题目链接:http://codeforces.com/problemset/problem/633/G 大意是一棵树两种操作,第一种是某一节点子树所有值+v,第二种问子树中节点模m出现了多少种m以内的质数. 第一种操作非常熟悉了,把每个节点dfs过程中的pre和post做出来,对序列做线段树.维护取模也不是问题.第二种操作,可以利用bitset记录质数出现情况.所以整个线段树需要维护bitset的信息. 对于某一个bitset x,如果子树所有值需要加y,则x=(x<<y)|(x>>…
题目链接:http://codeforces.com/problemset/problem/633/C 大意就是给个字典和一个字符串,求一个用字典中的单词恰好构成字符串的匹配. 比赛的时候是用AC自动机写的,就是对于trie中每一个节点,判断是否为终结点,以及当前字符所在位置p减去trie中这个节点的深度也即某一单词的长度l,判断dp[p-l]是否可以被构成,可以的话直接break并且标记当前dp值. 赛后想了想,其实直接一个trie就行了,每个单词才1000的长度,又想起来我去年给人讲过类似的…
B. A Trivial Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks f…
暴力 A - Ebony and Ivory import java.util.*; import java.io.*; public class Main { public static void main(String[] args) { Scanner cin = new Scanner (new BufferedInputStream (System.in)); int a = cin.nextInt (); int b = cin.nextInt (); int c = cin.nex…
B. A Trivial Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks f…
H. Fibonacci-ish II 题目连接: http://codeforces.com/contest/633/problem/H Description Yash is finally tired of computing the length of the longest Fibonacci-ish sequence. He now plays around with more complex things such as Fibonacci-ish potentials. Fibo…
E. Startup Funding 题目连接: http://codeforces.com/contest/633/problem/E Description An e-commerce startup pitches to the investors to get funding. They have been functional for n weeks now and also have a website! For each week they know the number of u…
G. Yash And Trees 题目连接: http://www.codeforces.com/contest/633/problem/G Description Yash loves playing with trees and gets especially excited when they have something to do with prime numbers. On his 20th birthday he was granted with a rooted tree of…
D. Fibonacci-ish 题目连接: http://www.codeforces.com/contest/633/problem/D Description Yash has recently learnt about the Fibonacci sequence and is very excited about it. He calls a sequence Fibonacci-ish if the sequence consists of at least two elements…
C. Spy Syndrome 2 题目连接: http://www.codeforces.com/contest/633/problem/C Description After observing the results of Spy Syndrome, Yash realised the errors of his ways. He now believes that a super spy such as Siddhant can't use a cipher as basic and a…
A. Ebony and Ivory 题目连接: http://www.codeforces.com/contest/633/problem/A Description Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of t…
D. Fibonacci-ish time limit per test 3 seconds memory limit per test 512 megabytes input standard input output standard output Yash has recently learnt about the Fibonacci sequence and is very excited about it. He calls a sequence Fibonacci-ish if th…
time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After observing the results of Spy Syndrome, Yash realised the errors of his ways. He now believes that a super spy such as Siddhant can't u…
time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, E…
还是dfs? 好像自己写的有锅 过不去 看了题解修改了才过qwq #include <cstdio> #include <algorithm> #include <cstring> #include <stack> #include <set> using namespace std; ],v[],n,m,k,ans; ],vis[]; ]; stack<int> S; void dfs(int u){ if(sons[u].size…
随便模拟下就过了qwq 然后忘了特判WA了QwQ #include <cstdio> #include <algorithm> #include <cstring> #include <set> #include <queue> using namespace std; ],v[],fir[],nxt[],cnt=,dep[],squ[]; ]; void addedge(int ui,int vi){ cnt++; u[cnt]=ui; v[c…
是一道水题 虽然看起来像是DP,但其实是贪心 扫一遍就A了 QwQ #include <cstdio> #include <algorithm> #include <cstring> #include <set> #include <map> using namespace std; ],b[],f[]; ],bx[]; int main(){ scanf("%d",&n); scanf(); scanf(); ;i&…
题目要求很简单,做法很粗暴 直接扫一遍即可 注意结果会爆int #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; ,n,s,a[]; int main(){ scanf("%I64d %I64d",&n,&s); ;i<=n;i++) scanf("%I64d&quo…
就是找一下规律 但是奈何昨天晚上脑子抽 推错了一项QwQ 然后重新一想 A掉了QwQ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; int main(){ int n; scanf("%d",&n); ,ans=,now=; while(i<n){ i+=now*; now*=;…
题目链接:点击打开链接 题意:给你n个数, 问最长的题目中定义的斐波那契数列.  思路:枚举開始的两个数, 由于最多找90次, 所以能够直接暴力, 用map去重.  注意, 该题卡的时间有点厉害啊. 用了两个map结果超时. 细节參见代码: #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<string> #include<v…
建trie树,刚好字符串是反向的,直接在原图上向前搜索就OK了……………… 可怜的我竟然用了RK来hash,在test67那里T了…… 贴个RK的 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <set> #include <vector> #define LL…
寻找树上最大权值和的两条不相交的路径. 树形DP题.挺难的,对于我…… 定义三个变量ma[MAXN], t[MAXN], sum[MAXN] 其中,ma[i]代表i子树中,最长的路径和 t[i]代表i子树中,用来维护已有一条路径,而且还有一条链从叶子节点到i,则可以从根节点i向上扩展.如下图,维护红色部分 sum[i]维护从某叶子节点到根节点i的最长路径. 转移方程可以看代码,很容易明白 #include <iostream> #include <cstdio> #include…
[链接]h在这里写链接 [题意] 给你n个数字; 让你在其中找出三个数字i,j,k(i<=j<=k); 使得p*a[i]+q*a[j]+r*a[k]最大; [题解] /*     有一个要求i<=j<=k;     也就是说系数之间的下标是有关系的.     枚举第一个位置的下标i;         则第二个人j         i<=j<n         枚举中间那个人的位置在哪.     求出前i个位置,第一个人能获得的最大值是多少     求出后n-i个位置,第…
[链接]h在这里写链接 [题意] 在这里写题意 [题解] /* Be careful. 二重循环枚举 */ [错的次数] 0 [反思] 在这了写反思 [代码] #include <bits/stdc++.h> using namespace std; int n; string s[110]; int main() { //freopen("F:\\rush.txt", "r", stdin); scanf("%d", &n)…
Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer m and asks for the number of positive integers n, such that the factorial of n ends with exactly m zeroes. Are you among those great programmer…
Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2)-C. Magic Grid-构造 [Problem Description] ​ 给你一个\(n\),构造一个\(n\times n\)的矩阵,使其满足任意一行,或一列的异或值相同.保证\(n\)能被\(4\)整除. [Solution] ​ 可以发现,从\(0\)开始,每\(4\)个连续数的异或值为\(0\),所以可以很容易使得每一行的异或值等于\(0\). ​ 列…
Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2)-D. Restore Permutation-构造+树状数组 [Problem Description] ​ 给你一个长度为\(n\)的数组,第\(i\)个元素\(s_i\)表示一个排列中第\(i\)个元素之前,并且小于\(p_i\)的元素的和.求出满足此条件的排列. [Solution] ​ 假设\(n=5\),\(s[]=\{0,0,3,7,3\}\).从后往前看…
Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2)-E. Let Them Slide-思维+数据结构 [Problem Description] ​ \(n\times w\)的方格中,每一行有\(cnt_i\)个数字,每一行的数字都连续的放在一起,但是可以任意的平移.问每一列的最大和为多少. [Solution] ​ 最直观的想法是对于每一列\(j\),计算出每一行中的一个区间最大值,此区间长度为\(len=w-c…