ZOJ 1334 Basically Speaking

最新推荐文章于 2016-07-25 20:15:00 发布

原创最新推荐文章于 2016-07-25 20:15:00 发布 · 635 阅读

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文章标签：

#prototype #input #output #numbers #features #manager

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本文介绍了一个用于实现不同数制间转换的程序设计方案。该程序能够处理从2到16进制之间的任意转换，并且能够在7位显示屏幕上正确展示转换后的结果。文章提供了完整的C语言代码实现，包括输入处理、进制转换逻辑以及输出格式化。

Basically Speaking

Time Limit: 2 Seconds Memory Limit: 65536 KB

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

题意：进制的转换

注意：还有用scanf(%s)时，空格不会被读进来

代码：

#include <stdio.h>
#include <string.h>
int main()
{
char str[30];
int b1,b2;
int inta[71];
int len;
int i,j;
inta['0']=0;
inta['1']=1;
inta['2']=2;
inta['3']=3;
inta['4']=4;
inta['5']=5;
inta['6']=6;
inta['7']=7;
inta['8']=8;
inta['9']=9;
inta['A']=10;
inta['B']=11;
inta['C']=12;
inta['D']=13;
inta['E']=14;
inta['F']=15;
while(scanf("%s %d %d",str,&b1,&b2)!=EOF) {
len=strlen(str);
int k=1;
int sum=0;
int m;
i=len-1;
while(i>=0)
{
sum+=inta[str[i]]*k;
k=k*b1;
i--;
}
i=0;
while(sum>0)
{
int temp=sum%b2;
if(temp>=10)
str[i++]=(char)(temp-10+'A');
else
str[i++]=(char)(temp+'0');
sum/=b2;
}
if(i>7)
{
printf(" ERROR\n");
}
else
{
for(m=0;m<7-i;m++)
printf(" ");
for(j=i-1;j>=0;j--)
printf("%c",str[j]);
printf("\n");
}
}
return 0;
}