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最新推荐文章于 2018-10-13 18:25:57 发布

xjbscut123456

最新推荐文章于 2018-10-13 18:25:57 发布

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文章标签： numbers string input each output pair

本文链接：https://blog.youkuaiyun.com/xjbscut123456/article/details/5398262

本文介绍了一个编程挑战——火星加法竞赛，参赛者需为22世纪火星数学竞赛编写程序，实现两个100位20进制数相加的功能。文章提供了C++代码示例，包括输入转换、加法运算及输出等关键步骤。

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Martian Addition

Time Limit: 1 Second Memory Limit: 32768 KB

In the 22nd Century, scientists have discovered intelligent residents live on the Mars. Martians are very fond of mathematics. Every year, they would hold an Arithmetic Contest on Mars (ACM). The task of the contest is to calculate the sum of two 100-digit numbers, and the winner is the one who uses least time. This year they also invite people on Earth to join the contest.
As the only delegate of Earth, you're sent to Mars to demonstrate the power of mankind. Fortunately you have taken your laptop computer with you which can help you do the job quickly. Now the remaining problem is only to write a short program to calculate the sum of 2 given numbers. However, before you begin to program, you remember that the Martians use a 20-based number system as they usually have 20 fingers.

Input:
You're given several pairs of Martian numbers, each number on a line.
Martian number consists of digits from 0 to 9, and lower case letters from a to j (lower case letters starting from a to present 10, 11, ..., 19).
The length of the given number is never greater than 100.

Output:
For each pair of numbers, write the sum of the 2 numbers in a single line.

Sample Input:

1234567890
abcdefghij
99999jjjjj
9999900001

Sample Output:

bdfi02467j
iiiij00000


#include <iostream>
#include <iomanip>
#include <string>
using namespace std;

void swap(char &s1,char &s2)
{
char temp;
temp=s1;
s1=s2;
s2=temp;
}
void convert(string& s)
{
int len=s.size();
for(int i=0,j=len-1;i<j;i++,j--)
  swap(s[i],s[j]);
}
int change(char c)
{
if(c>='0'&&c<='9')return c-'0';
else return c-'a'+10;
}
char rechange(int n)
{
if(n>=0&&n<=9)return (char)n+'0';
else return (char)n+'a'-10;
}
void add(string s1,string s2,char* s3)
{
int i;
convert(s1);
convert(s2);
int len1=s1.size();
int len2=s2.size();
int sum;
int tag=0;
for(i=0;i<len1&&i<len2;i++)
{
  sum=change(s1[i])+change(s2[i])+tag;
  if(sum>=20)s3[i]=rechange(sum-20),tag=1;
  else s3[i]=rechange(sum),tag=0;
}
if(i<len1)
{
  for(;i<len1;i++)
  {
   sum=change(s1[i])+tag;
   if(sum>=20)s3[i]=rechange(sum-20),tag=1;
   else s3[i]=rechange(sum),tag=0;
  }
}
if(i<len2)
{
  for(;i<len2;i++)
  {
   sum=change(s2[i])+tag;
   if(sum>=20)s3[i]=rechange(sum-20),tag=1;
   else s3[i]=rechange(sum),tag=0;
  }
}
if(tag)s3[i]=rechange(1),s3[i+1]='/0';
else s3[i]='/0';
string answer(s3);
convert(answer);
cout<<answer<<endl;
//cout<<s3[i]<<endl;
//for(int j=i-1;j>=0;j--)cout<<s3[j];
//cout<<endl;
}