Manthan, Codefest 16 A. Ebony and Ivory 水题
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 them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.
Input
The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output
Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.
Sample Input
4 6 15
Sample Output
No
Hint
题意
给你 a,b,c
问你能够使用若干个a和若干个b,恰好组成c
题解:
暴力枚举一个,然后O(1)算另外一个就好了
代码
#include<bits/stdc++.h>
using namespace std;
int main()
{
long long a,b,c;
cin>>a>>b>>c;
for(int i=0;i<=100000;i++)
{
long long res = c - a*i;
if(res==0)return puts("Yes");
if(res<0)break;
long long p = res/b;
if(p*b==res)
return puts("Yes");
}
return puts("No");
}
Manthan, Codefest 16 A. Ebony and Ivory 水题的更多相关文章
- Manthan, Codefest 16 -A Ebony and Ivory
time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standa ...
- Manthan, Codefest 16
暴力 A - Ebony and Ivory import java.util.*; import java.io.*; public class Main { public static void ...
- Manthan, Codefest 16 D. Fibonacci-ish
D. Fibonacci-ish time limit per test 3 seconds memory limit per test 512 megabytes input standard in ...
- Manthan, Codefest 16(B--A Trivial Problem)
B. A Trivial Problem time limit per test 2 seconds memory limit per test 256 megabytes input standar ...
- Manthan, Codefest 16 -C. Spy Syndrome 2
time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standa ...
- CF Manthan, Codefest 16 G. Yash And Trees 线段树+bitset
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 ...
- CF #Manthan, Codefest 16 C. Spy Syndrome 2 Trie
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 ...
- CF Manthan, Codefest 16 B. A Trivial Problem
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 ...
- Manthan, Codefest 16 H. Fibonacci-ish II 大力出奇迹 莫队 线段树 矩阵
H. Fibonacci-ish II 题目连接: http://codeforces.com/contest/633/problem/H Description Yash is finally ti ...
随机推荐
- python中的Queue模块
queue介绍 queue是python的标准库,俗称队列.可以直接import引用,在python2.x中,模块名为Queue.python3直接queue即可 在python中,多个线程之间的数据 ...
- 【Python学习笔记】Pandas库之DataFrame
1 简介 DataFrame是Python中Pandas库中的一种数据结构,它类似excel,是一种二维表. 或许说它可能有点像matlab的矩阵,但是matlab的矩阵只能放数值型值(当然matla ...
- python实战===用python调用jar包(原创)
一个困扰我很久的问题,今天终于解决了.用python调用jar包 很简单,但是网上的人就是乱转载.自己试都不试就转载,让我走了很多弯路 背景:python3.6 32位 + jre 32位 + ...
- 【Educationcal Codeforces Round 21】
这场edu我原本以为能清真一点…… 后来发现不仅是七题 还有各种奇奇怪怪的骚操作…… A. 随便枚举 #include<bits/stdc++.h> using namespace std ...
- 2017多校第9场 HDU 6162 Ch’s gift 树剖加主席树
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6162 题意:给出一棵树的链接方法,每个点都有一个数字,询问U->V节点经过所有路径中l < ...
- Struts2学习笔记04 之 拦截器
一.创建拦截器组件 1. 创建一个类,实现Interceptor接口,并实现intercept方法 2.注册拦截器 3.引用拦截器 二.拦截器栈 预置拦截器: 默认引用拦截器 拦截器调用顺序: Fil ...
- 从Java Future到Guava ListenableFuture实现异步调用
原文地址: http://blog.csdn.net/pistolove/article/details/51232004 Java Future 通过Executors可以创建不同类似的线程 ...
- graylog安装
官网:http://docs.graylog.org/en/2.4/pages/installation/os/centos.html Prerequisites Taking a minimal s ...
- 安装vmware+CentOS 7.4
安装步骤 选择第一个 按tab键 空格下一行 输入 红框内容 回车 注意事项 道路不通排查过程1.ip地址2.vmware 编辑-虚拟网络编辑器3.windows 服务 vmware相关服务 要开启 ...
- 二:Storm的配置项说明
配置项 配置说明 storm.zookeeper.servers ZooKeeper服务器列表 storm.zookeeper.port ZooKeeper连接端口 storm.local.dir s ...