Have meal

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 756    Accepted Submission(s): 507

Problem Description
I
have been in school for several years, so I have visited all messes
here. Now I have lost intersts in all of the foods. So when during the
meal time, I don’t know which mess I should go to. So I came up with a
solution.
There are 4 messes in our school, I number them from 0 to
3. Then I says “Big Bing Small Jiang, Point Who Is Who!”, when I say the
first word I point to the mess which is numbered 0, when I say the i-th
(i>1) word I point to the mess whose number is one larger than the
previous one. In case of the number of previous mess is 3, I will point
to 0 again. I will go to the mess which I point to last time. Thus in
this case I will go to the mess which is numbered 3. The following table
explains the course of my solution to this case.
 Word I say Mess id I point to Big 0 Bing 1 Small 2 Jiang 3 Point 0 Who 1 Is 2 Who 3
I
will go to university after several days, I have heard that there are
so many messes in it. So I will apply my solution again. Surpose there
are n messes which are numberd through 0 to n-1, and I will say m words.
When I say the first word I point to the mess which is numbered 0, when
I say the i-th (i>1) word I point to the mess whose number is one
larger than the previous one. In case of the number of previous mess is
n-1, I will point to 0 again. I will go to the mess which I point to
last time. So which mess will I point to?.
It is so time-consuming to count it through manual work. So I want you to write a program to help me. Would you help me?
 
Input
Multi test cases (about 10000), every case contain two integers n and m in a single line.

[Technical Specification]
1<=n, m<=100

 
Output
For each case, output the number of the mess which I should go to.
 
Sample Input
4 3
1 100
 
Sample Output
2
0
 
Source
 
题意:有n个菜(编号0-n-1),现在有一个人点菜,他点菜的规则是念m个单词,然后按照循环顺序数这n个菜,问最后会点到哪个菜??
#include<stdio.h>
#include<iostream>
#include<string.h>
#include <stdlib.h>
#include<math.h>
#include<algorithm>
#include <queue>
using namespace std;
typedef long long LL; int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF){
int t = m%n;
if(t==) printf("%d\n",n-);
else printf("%d\n",t-);
}
return ;
}

hdu 5158(水题)的更多相关文章

  1. HDU-1042-N!(Java大法好 &amp;&amp; HDU大数水题)

    N! Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Subm ...

  2. HDU 5391 水题。

    E - 5 Time Limit:1500MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Statu ...

  3. hdu 1544 水题

    水题 /* * Author : ben */ #include <cstdio> #include <cstdlib> #include <cstring> #i ...

  4. HDU排序水题

    1040水题; These days, I am thinking about a question, how can I get a problem as easy as A+B? It is fa ...

  5. hdu 2710 水题

    题意:判断一些数里有最大因子的数 水题,省赛即将临近,高效的代码风格需要养成,为了简化代码,以后可能会更多的使用宏定义,但是通常也只是快速拿下第一道水题,涨自信.大部分的代码还是普通的形式,实际上能简 ...

  6. Dijkstra算法---HDU 2544 水题(模板)

    /* 对于只会弗洛伊德的我,迪杰斯特拉有点不是很理解,后来发现这主要用于单源最短路,稍稍明白了点,不过还是很菜,这里只是用了邻接矩阵 套模板,对于邻接表暂时还,,,没做题,后续再更新.现将这题贴上,应 ...

  7. hdu 5162(水题)

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5162 题解:看了半天以为测试用例写错了.这题玩文字游戏.它问的是当前第i名是原数组中的第几个. #i ...

  8. hdu 3357 水题

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3357 #include <cstdio> #include <cmath> # ...

  9. hdu 5038 水题 可是题意坑

    http://acm.hdu.edu.cn/showproblem.php?pid=5038 就是求个众数  这个范围小 所以一个数组存是否存在的状态即可了 可是这句话真恶心  If not all ...

随机推荐

  1. Nodejs-异步操作

    1.阻塞 console.time('main');//代码计时器 //不断循环阻塞了代码的执行 for(var i=0;i<10000000;i++){ } console.timeEnd(' ...

  2. 这是我见过最厉害的--智能代码生成器、html+js+底层+sql全都有、瓦特平台

    1:直接上图.图片有点多.我就没全部上传了. (demo.使用方法.数据库bak)下载:http://pan.baidu.com/s/1ntE5bDn 起源: 之前有好多人问我代码生成器的源码.我发了 ...

  3. ASP NET Core 部署 IIS 和发布

    1. 微软官网原文链接: https://docs.microsoft.com/zh-cn/aspnet/core/host-and-deploy/iis/index?view=aspnetcore- ...

  4. 1043 Is It a Binary Search Tree (25 分)(二叉查找树)

    #include<bits/stdc++.h> using namespace std; typedef struct node; typedef node *tree; struct n ...

  5. Qt(1)

    Qt Qt开发图形界面软件,可以跨win.linux.mac平台.移动端,使用c++开发 Qt采用所见即所得的UI设计(UI设计和代码是联动的),GUI界面编辑信号和槽,由开发环境自动生成c++代码, ...

  6. Android记事本07

    昨天: activity横竖屏切换的生命周期 今天: Anr异常的原因和解决方案 遇到的问题: 无.

  7. 启动Tomcat时的常见问题及解决办法

    问题一:环境变量 1.检查jdk 验证jdk的配置,在运行-cmd中输入 java -version 即表示安装成功. 如果jdk没有问题,还需要配置两个环境变量.找到jdk和jre的路径,配置JAV ...

  8. RabbitMQ vhost 配置

    RabbitMQ vhost 配置 rabbitmqctl set_vhost_limits是用来定义虚拟主机限制的命令 配置最大连接限制 要限制vhost vhost_name中并发客户端连接的 总 ...

  9. P1194 买礼物

    题目描述 又到了一年一度的明明生日了,明明想要买B样东西,巧的是,这B样东西价格都是A元. 但是,商店老板说最近有促销活动,也就是: 如果你买了第I样东西,再买第J样,那么就可以只花K[I,J]元,更 ...

  10. POJ 2987 Firing | 最大权闭合团

    一个点带权的图,有一些指向关系,删掉一个点他指向的点也不能留下,问子图最大权值 题解: 这是最大权闭合团问题 闭合团:集合内所有点出边指向的点都在集合内 构图方法 1.S到权值为正的点,容量为权值 2 ...