codeforces Ebony and Ivory(水题)

时间:2022-09-24 07:41:59
A. Ebony and Ivory
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, 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 abc (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.

Examples
input
4 6 15
output
No
input
3 2 7
output
Yes
input
6 11 6
output
Yes
Note

In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.

题意:给你a  b  c三个数,问是否存在x  y使得a*x+b*y=c

数据不大直接暴力求解

#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD double
#define MAX 100100
#define mod 10007
#define dian 1.000000011
using namespace std;
int main()
{
int a,b,c,t,n,m,i,j;
while(scanf("%d%d%d",&a,&b,&c)!=EOF)
{
int flag=0;
for(i=0;i<=c/a;i++)
{
if((c-i*a)%b==0)
{
flag=1;
break;
}
}
if(flag) printf("Yes\n");
else printf("No\n");
}
return 0;
}

  

codeforces Ebony and Ivory(水题)的更多相关文章

  1. Manthan&comma; Codefest 16 A&period; Ebony and Ivory 水题

    A. Ebony and Ivory 题目连接: http://www.codeforces.com/contest/633/problem/A Description Dante is engage ...

  2. Codeforces Gym 100531G Grave 水题

    Problem G. Grave 题目连接: http://codeforces.com/gym/100531/attachments Description Gerard develops a Ha ...

  3. codeforces 706A A&period; Beru-taxi&lpar;水题&rpar;

    题目链接: A. Beru-taxi 题意: 问那个taxi到他的时间最短,水题; AC代码: #include <iostream> #include <cstdio> #i ...

  4. codeforces 569B B&period; Inventory&lpar;水题&rpar;

    题目链接: B. Inventory time limit per test 1 second memory limit per test 256 megabytes input standard i ...

  5. Codeforces 489A SwapSort (水题)

    A. SwapSort time limit per test 1 second memory limit per test 256 megabytes input standard input ou ...

  6. codeforces 688A A&period; Opponents&lpar;水题&rpar;

    题目链接: A. Opponents time limit per test 1 second memory limit per test 256 megabytes input standard i ...

  7. CodeForces 534B Covered Path &lpar;水题&rpar;

    题意:给定两个速度,一个一初速度,一个末速度,然后给定 t 秒时间,还每秒速度最多变化多少,让你求最长距离. 析:其实这个题很水的,看一遍就知道怎么做了,很明显就是先从末速度开始算起,然后倒着推. 代 ...

  8. Codeforces Gym 100286I iSharp 水题

    Problem I. iSharpTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/ ...

  9. CodeForces 705A(训练水题)

    题目链接:http://codeforces.com/problemset/problem/705/A 从第三个输出中可看出规律, I hate that I love that I hate it ...

随机推荐

  1. 【原】Mac下统计任意文件夹中代码行数的工具——cloc

    这里介绍一个Mac系统统计代码行数的工具cloc. 1.首先,安装homebrew,已安装的请跳过. 打开终端工具Terminal,输入下列命令.过程中会让你按RETURN键以及输入mac桌面密码,按 ...

  2. Javascript备忘复习笔记1

    一.字符串操作 1.大小写 var s = "hello"; undefined g = s.toUpperCase(); "HELLO" g; "H ...

  3. url截取判断(实现同级列表)

    <script> var dUrl=window.location.href; var cUrl=(dUrl.substring(0, dUrl.indexOf('list_'))); v ...

  4. SQL查询显示行号、随机查询、取指定行数据

    转自:walkingp 1.显示行号 如果数据没有删除的情况下主键与行号是一致的,但在删除某些数据,行号就与主键不一致了,这时需要查询行号就需要用新的方法,在SQL Server2005之前,需要使用 ...

  5. JAVA里的String、Timestamp、Date相互转换

    Timestamp转化为String: SimpleDateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");//定义 ...

  6. Enum基础

    enum ColorE {     RED, GREEN, BLUE; } public class GetEnumContent {     public static void main(Stri ...

  7. KMS服务器搭建

  8. 基于TCP协议的网络编程

    TCP通信协议是一种可靠的传输层协议,它在通信的两端各建立一个Socket,从而在通信的两端之间形成虚拟网络链路.一旦建立了虚拟的网络链路,两端的程序就可以通过虚拟链路进行通信.Java使用Socke ...

  9. Spark踩坑记——从RDD看集群调度

    [TOC] 前言 在Spark的使用中,性能的调优配置过程中,查阅了很多资料,之前自己总结过两篇小博文Spark踩坑记--初试和Spark踩坑记--数据库(Hbase+Mysql),第一篇概况的归纳了 ...

  10. C&num;中Dictionary的介绍

    关键字:C# Dictionary 字典 作者:txw1958原文:http://www.cnblogs.com/txw1958/archive/2012/11/07/csharp-dictionar ...