CodeForces - 415B Mashmokh and Tokens
Bimokh is Mashmokh's boss. For the following n days he decided to pay to his workers in a new way. At the beginning of each day he will give each worker a certain amount of tokens. Then at the end of each day each worker can give some of his tokens back to get a certain amount of money. The worker can save the rest of tokens but he can't use it in any other day to get more money. If a worker gives back w tokens then he'll get
dollars.
Mashmokh likes the tokens however he likes money more. That's why he wants to save as many tokens as possible so that the amount of money he gets is maximal possible each day. He has n numbers x1, x2, ..., xn. Number xi is the number of tokens given to each worker on the i-th day. Help him calculate for each of n days the number of tokens he can save.
Input
The first line of input contains three space-separated integers n, a, b (1 ≤ n ≤ 105; 1 ≤ a, b ≤ 109). The second line of input contains n space-separated integers x1, x2, ..., xn (1 ≤ xi ≤ 109).
Output
Output n space-separated integers. The i-th of them is the number of tokens Mashmokh can save on the i-th day.
Example
Input5 1 4
12 6 11 9 1Output0 2 3 1 1Input3 1 2
1 2 3Output1 0 1Input1 1 1
1Output0
1 #include<iostream>
2 #include<stdio.h>
3 using namespace std;
4 long long int num[100010];
5 int main()
6 {
7 int n;
8 while(cin>>n)
9 {
10 long long int a,b;
11 scanf("%lld%lld",&a,&b);
12 for(int i=0;i<n;i++)
13 cin>>num[i];
14 long long int ans;
15 for(int i=0;i<n;i++)
16 {
17 ans=((num[i]*a)%b)/a;
18 printf("%lld ",ans);
19 }
20 cout<<endl;
21 }
22 return 0;
23 }
CodeForces - 415B Mashmokh and Tokens的更多相关文章
- Codefroces 415B Mashmokh and Tokens
B. Mashmokh and Tokens time limit per test 1 second memory limit per test 256 megabytes input standa ...
- codeforces 414D Mashmokh and Water Tanks
codeforces 414D Mashmokh and Water Tanks 题意 题解 \(a_i\):第 \(i\) 层的结点个数. \(b_i\):第 \(i\) 层初始有水的结点个数. 如 ...
- Mashmokh and Tokens
Codeforces Round #240 (Div. 2) B;http://codeforces.com/problemset/problem/415/B 题意:老板一天发x张代币券,员工能用它来 ...
- Codeforces 414B Mashmokh and ACM
http://codeforces.com/problemset/problem/414/B 题目大意: 题意:一个序列B1,B2...Bl如果是好的,必须满足Bi | Bi + 1(a | b 代表 ...
- Codeforces 414C Mashmokh and Reverse Operation
题意:给你2^n个数,每次操作将其分成2^k份,对于每一份内部的数进行翻转,每次操作完后输出操作后的2^n个数的逆序数. 解法:2^n个数,可以联想到建立一棵二叉树的东西,比如 2,1,4,3就可以 ...
- codeforces D.Mashmokh and ACM
题意:给你n和k,然后找出b1, b2, ..., bl(1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n),并且对所有的bi+1%bi==0,问有多少这样的序列? 思路:dp[i][j] 表示长 ...
- codeforces C. Mashmokh and Numbers
题意:给你n和k,然后让你找出n个数使得gcd(a1,a2)+gcd(a3,a4)+......的和等于k: 思路:如果n为奇数,让前n-3个数的相邻两个数都为1,n-2和n-1两个数gcd为k-an ...
- CodeForces 415D Mashmokh and ACM
$dp$. 记$dp[i][j]$表示已经放了$i$个数字,并且第$i$个数字放了$j$的方案数.那么$dp[i][j] = \sum\limits_{k|j}^{} {dp[i - 1][k]}$ ...
- @codeforces - 414E@ Mashmokh's Designed Problem
目录 @description@ @solution@ @accepted code@ @details@ @description@ 给定一棵 n 个点的树,每个点的儿子是有序的. 现给定 m 次操 ...
- Codeforces Round #240 (Div. 2)(A -- D)
点我看题目 A. Mashmokh and Lights time limit per test:1 secondmemory limit per test:256 megabytesinput:st ...
随机推荐
- 《最新出炉》系列入门篇-Python+Playwright自动化测试-15-playwright处理浏览器多窗口切换
1.简介 浏览器多窗口的切换问题相比大家不会陌生吧,之前宏哥在java+selenium系列文章中就有介绍过.大致步骤就是:使用selenium进行浏览器的多个窗口切换测试,如果我们打开了多个网页,进 ...
- 探析ElasticSearch Kibana在测试工作中的实践应用
一. 为什么使用ES Kibana 离线数据测试中最重要的就是数据验证,一部分需要测试es存储数据的正确性,另一部分就需要验证接口从es取值逻辑的正确性.而为了验证es取值逻辑的正确性,就需要用到Ki ...
- 「acmhdu - 6314」Matrix
link. 首先将问题弱化为 1-d,我们待定容斥系数 \(f_i\),可以写出答案的式子:\(\sum\limits_{i=a}^nf_i\binom{n}{i}2^{n-i}\).解释就是,我们想 ...
- EarthChat SignalR原理讲解
SignalR原理讲解 SignalR是什么? SignalR 是 Microsoft 开发的一个库,用于 ASP.NET 开发人员实现实时 web 功能.这意味着服务端代码可以实时地推送内容到连接的 ...
- Sentinel系列之SlotChain、NodeSelectorSlot、ClusterBuilderSlot分析
本文基于Sentinel 1.8.6版本分析 1. SlotChain 我们从入口com.alibaba.csp.sentinel.SphU#entry(java.lang.String) 开始分析. ...
- 【IOC,AOP】spring的基础概念
IOC 控制反转 对象的创建控制权转交给外部实体,就是控制反转.外部实体便是IOC容器.其实就是以前创建java对象都是我们new一下,现在我们可以把这个new交给IOC容器来做,new出来的对象也会 ...
- 普冉PY32系列(八) GPIO模拟和硬件SPI方式驱动无线收发芯片XN297LBW
目录 普冉PY32系列(一) PY32F0系列32位Cortex M0+ MCU简介 普冉PY32系列(二) Ubuntu GCC Toolchain和VSCode开发环境 普冉PY32系列(三) P ...
- Unity - EditorWindow 折叠树显示(IMGUI)
仅适用于2018之前的版本,有UIElements或者UIWidgets的最好用新的 基本实现 树节点 public interface ITreeNode { ITreeNode Parent { ...
- animate.css 动画种类(详细)
作者:WangMin 格言:努力做好自己喜欢的每一件事 以下为各种动画类型包含的不同动画效果类,仅供参考.具体可查看animate.css 官网. bounce 弹跳 2. flash 闪烁 3. p ...
- Python 潮流周刊#27:应该如何处理程序的错误?
你好,我是猫哥.这里每周分享优质的 Python.AI 及通用技术内容,大部分为英文.本周刊开源,欢迎投稿.另有电报频道作为副刊,补充发布更加丰富的资讯. 产品推荐 Walles.AI 是一款适用于所 ...