Martian Addition
Time Limit: 2 Seconds Memory Limit: 65536 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
Source: Zhejiang University Local Contest 2002, Preliminary
解题思路：
求两个20进制的数的和，用数组模拟相加就能够。
代码：
#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#include <map>
using namespace std;
map<int,char>change;char num[105];
int add1[105];
int add2[105];
int ans[105];
int jin[105];void prepare()
{
change[0]='0';change[1]='1';change[2]='2';change[3]='3';
change[4]='4';change[5]='5';change[6]='6';change[7]='7';
change[8]='8';change[9]='9';change[10]='a';change[11]='b';
change[12]='c';change[13]='d';change[14]='e';change[15]='f';
change[16]='g';change[17]='h';change[18]='i';change[19]='j';
}int main()
{
prepare();
while(cin>>num)
{
memset(add1,0,sizeof(add1));
memset(add2,0,sizeof(add2));
memset(jin,0,sizeof(jin));
int len1=strlen(num);
int t=0;
for(int i=len1-1;i>=0;--i)
{
if(!(num[i]>='0'&&num[i]<='9'))
add1[t++]=num[i]-87;
else
add1[t++]=num[i]-'0';
}
cin>>num;
int len2=strlen(num);
t=0;
for(int i=len2-1;i>=0;--i)
{
if(!(num[i]>='0'&&num[i]<='9'))
add2[t++]=num[i]-87;
else
add2[t++]=num[i]-'0';
}
//输入处理完成
//for(int i=0;i<len2;i++)
// cout<<add2[i];
//cout<<endl;int len=max(len1,len2);
for(int i=0;i<=len;i++)
{
int temp=add1[i]+add2[i]+jin[i];
if(temp>=40)
{
ans[i]=temp-40;
jin[i+1]=2;
}
else if(temp>=20)
{
ans[i]=temp-20;
jin[i+1]=1;
}
else
ans[i]=temp;
}
if(ans[len])
cout<<change[ans[len]];
for(int i=len-1;i>=0;i--)
cout<<change[ans[i]];
cout<<endl;
}
return 0;
}
