Problem G: If We Were a Child Again
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 18 Solved: 14
[Submit][Status][Web Board]
Description
The Problem

The first project for the poor student was to make a calculator that can just perform the basic arithmetic operations.

But like many other university students he doesn’t like to do any project by himself. He just wants to collect programs from here and there. As you are a friend of him, he asks you to write the program. But, you are also intelligent enough to tackle this kind of people. You agreed to write only the (integer) division and mod (% in C/C++) operations for him.

Input
Input is a sequence of lines. Each line will contain an input number. One or more spaces. A sign (division or mod). Again spaces. And another input number. Both the input numbers are non-negative integer. The first one may be arbitrarily long. The second number n will be in the range (0 < n < 231).

Output
A line for each input, each containing an integer. See the sample input and output. Output should not contain any extra space.

Sample Input
110 / 100
99 % 10
2147483647 / 2147483647
2147483646 % 2147483647
Sample Output
1
9
1
2147483646

my answer:

#include<iostream>
#include<cstring>
#include<string>
using namespace std;
int main()
{
char a[];
int t[];
int sum,b,i;
char c;
while(cin>>a)
{
cin>>c;
cin>>b;
int t1=strlen(a);
sum=a[]-''; for( i=;i!=t1;i++){
t[i]=sum/b;
sum=sum%b;
if(sum<b&&i<t1-){
sum=(sum*+a[i+]-''); }
}
if(c=='%')
cout<<sum<<endl;
else{
int start=;
while(!t[start])start++;
if(start==i)cout<<<<endl;
else {
for(int j=start;j<i;j++)
cout<<t[j];
cout<<endl;
}
}
}
return ;
}

Problem G: If We Were a Child Again的更多相关文章

  1. 实验9:Problem G: 克隆人来了!

    想要输出""的话: cout<<"A person whose name is \""<<name<<" ...

  2. 实验12:Problem G: 强悍的矩阵运算来了

    这个题目主要是乘法运算符的重载,卡了我好久,矩阵的乘法用3个嵌套的for循环进行,要分清楚矩阵的乘法结果是第一个矩阵的行,第二个矩阵的列所组成的矩阵. 重载+,*运算符时,可以在参数列表中传两个矩阵引 ...

  3. 烟大 Contest1024 - 《挑战编程》第一章:入门 Problem G: Check The Check(模拟国际象棋)

    Problem G: Check The Check Time Limit: 1 Sec  Memory Limit: 64 MBSubmit: 10  Solved: 3[Submit][Statu ...

  4. The Ninth Hunan Collegiate Programming Contest (2013) Problem G

    Problem G Good Teacher I want to be a good teacher, so at least I need to remember all the student n ...

  5. 【贪心+中位数】【新生赛3 1007题】 Problem G (K)

    Problem G Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Sub ...

  6. Problem G: Keywords Search

    Problem G: Keywords SearchTime Limit: 1 Sec Memory Limit: 128 MBSubmit: 10 Solved: 6[Submit][Status] ...

  7. BZOJ4977 八月月赛 Problem G 跳伞求生 set 贪心

    欢迎访问~原文出处——博客园-zhouzhendong 去博客园看该题解 题目传送门 - BZOJ4977 - 八月月赛 Problem G 题意 小明组建了一支由n名玩家组成的战队,编号依次为1到n ...

  8. Western Subregional of NEERC, Minsk, Wednesday, November 4, 2015 Problem G. k-palindrome dp

    Problem G. k-palindrome 题目连接: http://opentrains.snarknews.info/~ejudge/team.cgi?SID=c75360ed7f2c7022 ...

  9. ZOJ 4010 Neighboring Characters(ZOJ Monthly, March 2018 Problem G,字符串匹配)

    题目链接  ZOJ Monthly, March 2018 Problem G 题意  给定一个字符串.现在求一个下标范围$[0, n - 1]$的$01$序列$f$.$f[x] = 1$表示存在一种 ...

随机推荐

  1. Igor In the Museum(搜搜搜151515151515******************************************************1515151515151515151515)

    D. Igor In the Museum time limit per test 1 second memory limit per test 256 megabytes input standar ...

  2. swift 用协议实现代理传值功能

    1.功能简介 RootViewController中用个lable和一个按钮,点击按钮跳转到模态窗口.在模态窗口中有个TextField和一个按钮,输入文字点击关闭模态按钮后跳转到RootViewCo ...

  3. 使用Vitamio打造自己的Android万能播放器(1)——准备

    前言 虽然Android已经内置了VideoView组件和MediaPlayer类来支持开发视频播放器,但支持格式.性能等各方面都十分有限,这里与大家一起利用免费的Vitamio来打造属于自己的And ...

  4. iOS极光推送的基本使用

    昨天花了一下午的时间研究了下极光推送,也前也是没做过,不知道从何下手!才开始的时候一看官方的SDK感觉好难,不过经过一系列的捣鼓之后,手机收到了推送信息,感觉其实并没有那么难! 1.配置开发证书(得有 ...

  5. trie tree(字典树)

    hihocoder题目(http://hihocoder.com/problemset):#1014 trie树 #include <iostream> using namespace s ...

  6. python 学习笔记 9 -- Python强大的自省简析

    1. 什么是自省? 自省就是自我评价.自我反省.自我批评.自我调控和自我教育,是孔子提出的一种自我道德修养的方法.他说:“见贤思齐焉,见不贤而内自省也.”(<论语·里仁>)当然,我们今天不 ...

  7. 勉強すべきURL

    http://www.atmarkit.co.jp/ait/articles/1403/19/news034_2.html http://webdesignerwork.jp/web/responsi ...

  8. xampp环境安装swoole

    手动编译php运行环境经常遇到函数库依赖的问题,这个错误搞定了,又蹦出来那个错误,很棘手,为了快速搭建一个swoole开发环境,于是另辟蹊径,直接下载安装xampp for linux,然后在用xam ...

  9. 基于Socket的UDP和TCP编程介绍

    一.概述 TCP(传输控制协议)和UDP(用户数据报协议是网络体系结构TCP/IP模型中传输层一层中的两个不同的通信协议. TCP:传输控制协议,一种面向连接的协议,给用户进程提供可靠的全双工的字节流 ...

  10. surface 其实是UEFI与BIOS并存,借用官网的进入方法(少有更改)

    surface 其实是UEFI与BIOS并存,借用官网的进入方法(少有更改) 第一种: 1.       Swipe in from the right edge of the screen, and ...