E. Jzzhu and Apples time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has picked n apples from his big apple tree. All the apples are numbered from 1 to n. Now he wants to sell them to…

D. Jzzhu and Cities time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Jzzhu is the president of country A. There are n cities numbered from 1 to n in his country. City 1 is the capital of A.…

A. Jzzhu and Children time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are n children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from…

B. Jzzhu and Sequences time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has invented a kind of sequences, they meet the following property: You are given x and y, please calculate fn m…

今天老师(orz sansirowaltz)让我们做了很久之前的一场Codeforces Round #257 (Div. 1),这里给出A~C的题解,对应DIV2的C~E. A.Jzzhu and Chocolate time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has a big rectangular cho…

题目地址 先简单的总结一下这次CF,前两道题非常的水,可是第一题又是因为自己想的不够周到而被Hack了一次(或许也应该感谢这个hack我的人,使我没有最后在赛后测试中WA).做到C题时看到题目情况非常复杂,明显超出自己现在水平,卡了很久也没有好好做题.等了很长时间才开始看D题,而这时信心已经严重不足.赛后再看D题才发现自己比赛时理解错了题意,致使误以为题目非常复杂.而实际上,这个过程分析起来是非常容易的. 对于三种棱长度分别为a,b,c的石头,r最大为min(a,b,c)/2;. 如果两个石头要…

这场比赛的出题人挺有意思,全部magic成了青色. 还有题目中的图片特别有趣. 晚上没打,开virtual contest打的,就会前三道,我太菜了. 最后看着题解补了第四道. 比赛传送门 A. Angry Students 题目大意:有t队学生,每个学生有两种状态,生气(A)或不生气(P).(话说为什么生气的戴着圣诞帽哇)所有生气的人都会往前一个人丢雪球,被丢到的人也会变得生气,也会丢雪球.问你每队人中最后一个学生变得生气的时刻. 这题就是统计最长的连续的'P'当然前提是左边有生气的人. 代码…

Codeforces Round #713 (Div. 3) Editorial 记录一下自己写的前二题本人比较菜 A. Spy Detected! You are given an array a consisting of n (n≥3) positive integers. It is known that in this array, all the numbers except one are the same (for example, in the array [4,11,4,4]…

题目网址:http://codeforces.com/contest/1154/problem/ 题目意思:就是给你四个数,这四个数是a+b,a+c,b+c,a+b+c,次序未知要反求出a,b,c,d 题解:显然先求出这四个数中的最大数,然后分别减去其他三个数,即得a,b,c. #include<bits/stdc++.h> using namespace std; ]; int main() { ;i<=;i++) cin>>a[i]; sort(a+,a++); prin…

题目链接:http://codeforces.com/problemset/problem/450/B B. Jzzhu and Sequences time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has invented a kind of sequences, they meet the following pr…

Problem A A. Jzzhu and Children time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are n children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the chi…

D. Dynamic Problem Scoring time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vasya and Petya take part in a Codeforces round. The round lasts for two hours and contains five problems. For th…

A - Jzzhu and Children 找到最大的ceil(ai/m)即可 #include <iostream> #include <cmath> using namespace std; int main(){ int n,m; cin >> n >> m; ; ; ; i < n; ++ i){ cin >> a; if(maxv <= ceil(a/m)){ maxv = ceil(a/m); maxIdx = i+;…

题目链接:http://codeforces.com/problemset/problem/450/A ---------------------------------------------------------------------------------------------------------------------------------------------------------- 欢迎光临天资小屋:http://user.qzone.qq.com/593830943…

题目链接:http://codeforces.com/problemset/problem/450/B 题意很好懂,矩阵快速幂模版题. /* | 1, -1 | | fn | | 1, 0 | | fn-1 | */ #include <iostream> #include <cstdio> #include <cstring> using namespace std; typedef __int64 LL; LL mod = 1e9 + ; struct data {…

主题链接:http://codeforces.com/problemset/problem/449/A ---------------------------------------------------------------------------------------------------------------------------------------------------------- 欢迎光临天资小屋:http://user.qzone.qq.com/593830943…

题目链接:http://codeforces.com/problemset/problem/449/C 给你n个数,从1到n.然后从这些数中挑选出不互质的数对最多有多少对. 先是素数筛,显然2的倍数的个数是最多的,所以最后处理.然后处理3,5,7,11...的倍数的数,之前已经挑过的就不能再选了.要是一个素数p的倍数个数是奇数,就把2*p给2 的倍数.这样可以满足p倍数搭配的对数是最优的.最后处理2的倍数就行了. #include <bits/stdc++.h> using namespace…

Jzzhu has invented a kind of sequences, they meet the following property: You are given x and y, please calculate fn modulo 1000000007 (109 + 7). Input The first line contains two integers x and y (|x|, |y| ≤ 109). The second line contains a single i…

D - Jzzhu and Numbers 这个容斥没想出来... 我好菜啊.. f[ S ] 表示若干个数 & 的值 & S == S得 方案数, 然后用这个去容斥. 求f[ S ] 需要用SOSdp #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define…

C. Jzzhu and Chocolate time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k time…

A. Jzzhu and Children time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are n children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from…

题意:n个城市,中间有m条道路(双向),再给出k条铁路,铁路直接从点1到点v,现在要拆掉一些铁路,在保证不影响每个点的最短距离(距离1)不变的情况下,问最多能删除多少条铁路 分析:先求一次最短路,铁路的权值大于该点最短距离的显然可以删去,否则将该条边加入图中,再求最短路,记录每个点的前一个点,然后又枚举铁路,已经删去的就不用处理了,如果铁路权值大于该点最短距离又可以删去,权值相等时,该点的前一个点如果不为1,则这个点可以由其他路到达,这条铁路又可以删去. 由于本题中边比较多,最多可以有8x10^…

题目地址:http://codeforces.com/contest/474/problem/E 第一次遇到这样的用线段树来维护DP的题目.ASC中也遇到过,当时也非常自然的想到了线段树维护DP,可是那题有简单方法,于是就没写.这次最终写出来了.. 这题的DP思想跟求最长上升子序列的思想是一样的.仅仅只是这里的找前面最大值时会超时,所以能够用线段树来维护这个最大值,然后因为还要输出路径,所以要用线段树再来维护一个每一个数在序列中所在的位置信息. 手残了好多地方,最终调试出来了... 代码例如以下…

A. Sasha and Sticks 题目链接:http://codeforces.com/contest/832/problem/A 题目意思:n个棍,双方每次取k个,取得多次数的人获胜,Sasha先取,问Sasha是否可以取胜. 代码: //Author: xiaowuga #include <iostream> #include <algorithm> #include <set> #include <vector> #include <que…

B解题报告 算是规律题吧,,,x y z -x -y -z 注意的是假设数是小于0,要先对负数求模再加模再求模,不能直接加mod,可能还是负数 给我的戳代码跪了,,. #include <iostream> #include <cstring> #include <cstdio> using namespace std; long long x,y,z; long long n; int main() { cin>>x>>y; cin>&g…

A. Circle of Students      题目:https://codeforces.com/contest/1203/problem/A 题意:一堆人坐成一个环,问能否按逆时针或者顺时针正好是 1-n的顺序 思路:水题,把数组开两倍,或者标记当前位置都可以 #include<bits/stdc++.h> #define maxn 100005 #define mod 1000000007 using namespace std; typedef long long ll; int…

小结: A,B,F 切,C 没写 1ll 对照样例才发现,E,G 对照样例过,D 对照样例+看了其他人代码(主要急于看后面的题,能调出来的但偷懒了. CF1674A Number Transformation 考虑若 \(y\) 不能整除 \(x\) 则无解,否则一定存在一组解 \(a=1,b=y\div x\). #include<bits/stdc++.h> #define IOS ios::sync_with_stdio(false) #define TIE cin.tie(0),cou…

C.Journey 读错题目了...不是无向图,结果建错图了(喵第4样例是变成无向就会有环的那种图) 并且这题因为要求路径点尽可能多 其实可以规约为限定路径长的拓扑排序,不一定要用最短路做 #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm>…

Socks Problem Description: Arseniy is already grown-up and independent. His mother decided to leave him alone for m days and left on a vacation. She have prepared a lot of food, left some money and washed all Arseniy's clothes. Ten minutes before her…

题意: 这道英文题的题意稍稍有点复杂. 找长度为n的数字序列有多少种.这个序列可以分为n/k段,每段k个数字.k个数可以变成一个十进制的数Xi.要求对这每n/k个数,剔除Xi可被ai整除的情况,剔除X的第一个数(包括前导0)是bi的情况.问剩下的组合有多少种. 思路: 这题我是一波三折的.首先也没有考虑很多,看着可以暴力模拟过程,我就直接开始敲了,几个for循环敲出来,再把bug调一调和特殊情况考虑考虑,交了之后开始TLE,这时候意识到复杂度太大了,于是开始优化,做了(b[i])*(mmax/1…