http://acm.hdu.edu.cn/showproblem.php?pid=1335

Basically Speaking

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1885    Accepted Submission(s): 729

Problem Description
The Really Neato Calculator Company, Inc. has recently hired your team to help design their Super Neato Model I calculator. As a computer scientist you suggested to the company that it would be neato if this new calculator could convert among number bases. The company thought this was a stupendous idea and has asked your team to come up with the prototype program for doing base conversion. The project manager of the Super Neato Model I calculator has informed you that the calculator will have the following neato features: It will have a 7-digit display.
Its buttons will include the capital letters A through F in addition to the digits 0 through 9.
It will support bases 2 through 16.
 
Input
The input for your prototype program will consist of one base conversion per line. There will be three numbers per line. The first number will be the number in the base you are converting from. The second number is the base you are converting from. The third number is the base you are converting to. There will be one or more blanks surrounding (on either side of) the numbers. There are several lines of input and your program should continue to read until the end of file is reached.
 
Output
The output will only be the converted number as it would appear on the display of the calculator. The number should be right justified in the 7-digit display. If the number is to large to appear on the display, then print "ERROR'' (without the quotes) right justified in the display.
 
Sample Input
1111000 2 10
1111000 2 16
2102101 3 10
2102101 3 15
12312 4 2
1A 15 2
1234567 10 16
ABCD 16 15
 
Sample Output
   120
    78
   1765
   7CA
  ERROR
  11001
  12D687
  D071
 
Source
 
Recommend
Ignatius.L

 #include<stdio.h>
#include<string.h>
int cmp(int x,int y)
{
int s=,i;
for(i=;i<=y;i++)
s=s*x;
return s;
}
int main()
{
int t,n,s,l,i,r,j,k,p,q;
char str[];
int a[],b[];
while(~scanf("%s",str))
{
s=j=;
scanf("%d%d",&t,&n);
l=strlen(str);
for(i=;i<l;i++)
{
if(str[i]>=''&&str[i]<='')
a[j++]=str[i]-'';
else
a[j++]=str[i]-;
}
k=;
for(i=j-;i>=;i--)
{
r=a[i]*cmp(t,k);
s+=r;
k++;
}
q=;
while(s!=)
{
p=s%n;
b[q++]=p;
s=s/n;
}
for(j=;j<-q;j++)
{
if(q>)
break;
printf(" ");
}
for(i=q-;i>=;i--)
{
if(q>)
{
printf(" ERROR");
break;
}
if(b[i]==)
printf("A");
if(b[i]==)
printf("B");
if(b[i]==)
printf("C");
if(b[i]==)
printf("D");
if(b[i]==)
printf("E");
if(b[i]==)
printf("F");
if(b[i]>=&&b[i]<=)
printf("%d",b[i]);
}
printf("\n");
}
return ;
}
 
 

HDU-1335 Basically Speaking的更多相关文章

  1. HDU 1335 Basically Speaking(进制转换)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1335 Problem Description The Really Neato Calculator ...

  2. HDOJ 1335 Basically Speaking(进制转换)

    Problem Description The Really Neato Calculator Company, Inc. has recently hired your team to help d ...

  3. Basically Speaking

    Basically Speaking Time Limit: 2 Sec  Memory Limit: 200 MB Submit: 19  Solved: 11 [Submit][Status][W ...

  4. HDU 1314 Numerically Speaking(大数加减乘除+另类二十六进制互相转换)

    原题代号:HDU 1314 原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1314 Numerically Speaking Time Limit: 2 ...

  5. ZOJ Problem Set - 1334 Basically Speaking ac代码及总结

    这道题目不难,是一道简单的进制转换问题,但是发现了自己两个遗漏的知识点: 1.关于scanf (1)scanf函数在输入时是以回车或者空格作为一次输入的结束 (2)scanf函数在输入字符串的过程中是 ...

  6. HDU——PKU题目分类

    HDU 模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 ...

  7. POJ题目分类(按初级\中级\高级等分类,有助于大家根据个人情况学习)

    本文来自:http://www.cppblog.com/snowshine09/archive/2011/08/02/152272.spx 多版本的POJ分类 流传最广的一种分类: 初期: 一.基本算 ...

  8. POJ题目细究

    acm之pku题目分类 对ACM有兴趣的同学们可以看看 DP:  1011   NTA                 简单题  1013   Great Equipment     简单题  102 ...

  9. 【转】POJ百道水题列表

    以下是poj百道水题,新手可以考虑从这里刷起 搜索1002 Fire Net1004 Anagrams by Stack1005 Jugs1008 Gnome Tetravex1091 Knight ...

随机推荐

  1. java连接远程服务器之manyged-ssh2 (windows和linux)

    一.所需要的jar包 需要借助Ganymed SSH的jar包:  ganymed-ssh2-262.jar 下载地址: http://www.ganymed.ethz.ch/ssh2/ API详情: ...

  2. Codevs 1800 假面舞会 2008年NOI全国竞赛

    1800 假面舞会 2008年NOI全国竞赛 时间限制: 1 s 空间限制: 128000 KB 题目等级 : 大师 Master 题目描述 Description 一年一度的假面舞会又开始了,栋栋也 ...

  3. windows phone 8 设置锁屏背景

    本来想研究一下 利用闪光灯实现手电筒的代码,发现不是简答设置FlashMode属性可以解决问题的,ms也没有提供api,无意瞄了一眼侧边栏的文章列表,发现了设置屏幕锁屏背景的实现,手一抖点进去了.还算 ...

  4. php安装过程中遇到的需要安装的问题

    http://www.cnblogs.com/kristain/articles/3809243.html     借鉴php安装错误 2013-01-04 19:16:49 分类: 系统运维 环境: ...

  5. 判断JS对象是否拥有某属性的方法举例

    判断JS对象是否拥有某属性 JS是否拥有某属性的判断方法,这里提供两种方式,供大家参考. 1.in 运算符 var obj = {name:'jack'}; alert('name' in obj); ...

  6. WORDPRESS插件开发(二)HELLO WORLD改进版

    在上一篇文章中WORDPRESS插件开发(一)HELLO WORLD,演示了Hello World的最简单实现,只是在每篇文章的后面加入Hello World字符,而且字符也是写死的. 如果用户需要自 ...

  7. PHP实现多web服务器共享SESSION数据-session数据写入mysql数据库

    http://www.php100.com/html/webkaifa/PHP/PHPyingyong/2010/0822/5276.html http://hi.baidu.com/lei_com/ ...

  8. Poco之ftp获取文件列表以及下载文件

    #include <iostream>#include <string>#include <vector>#include <algorithm>#in ...

  9. 快速生成json实体类

    读取一个json文件,并与实体相对应: static void Main(string[] args) { string json = ""; FileStream fs = ne ...

  10. Django中国|Django中文社区——python、django爱好者交流社区

    Django中国致力于成为Python和Django框架等技术的中文开发者学习交流平台. 内容涵盖python教程.python基础.Django教程.python入门.web.py教程.linux教 ...