ZOJ Problem Set - 1205

Martian Addition

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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  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

解题报告：

用两个数组模拟两个数，进行加减。

#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()                        //数组记录1~9，a~j； 
{  
    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)         //输入第一组数 
    {  
        memset(add1,0,sizeof(add1));      //add1[i]记录第一组数 
        memset(add2,0,sizeof(add2));      //add2[i]记录第二组数 
        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;     //ASCII码 a~j为97~106     10≤num[i]-87≤19        
            else  
                add1[t++]=num[i]-'0';     //‘0’≤num[i]≤‘9’ 
        }  
        cin>>num;            //输入第二组数 
        int len2=strlen(num);    
        t=0;  
        for(int i=len2-1;i>=0;--i)  
        {  
            if(!(num[i]>='0'&&num[i]<='9'))    //ASCII码 a~j为97~106     10≤num[i]-87≤19
                add2[t++]=num[i]-87;  
            else  
                add2[t++]=num[i]-'0';          // ‘0’≤num[i]≤‘9’
        }//输入处理完毕  

        
		int len=max(len1,len2);     
        for(int i=0;i<=len;i++)  
        {  
            int temp=add1[i]+add2[i]+jin[i];    //两个数每一位开始相加    jin[i]上一位数相加大于20进的位 
            if(temp>=40)                       
            {  
                ans[i]=temp-40;                //两个数每一位相加大于40进2；记录余留下的小于20的数； 
                jin[i+1]=2;                    //两个数每一位相加大于40进2,20进制 
            }  
            else if(temp>=20)  
            {  
                ans[i]=temp-20;                //两个数每一位相加大于20进1；记录余留下的小于20的数；
                jin[i+1]=1;                   //两个数每一位相加大于20进1,20进制 
            }  
            else  
                ans[i]=temp;                 //两个数每位相加小于20不进位，直接保留下来 
        }  
        if(ans[len])       //ans[len]最高位为0则跳过，不为零输出最高位，因为会出现原先的两个数的最高位相加大于二十进一 
            cout<<change[ans[len]];
        for(int i=len-1;i>=0;i--)  
            cout<<change[ans[i]];         //输出每一位数 
        cout<<endl;  
    }  
    return 0;  
}