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[ACM] ZOJ Martian Addition (20進制的兩個大數相加）

編輯：C++入門知識

[ACM] ZOJ Martian Addition (20進制的兩個大數相加）

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

解題思路：

求兩個20進制的數的和，用數組模擬相加就可以。

代碼：

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
mapchange;

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';
		}
		//輸入處理完畢
		//for(int i=0;i=20)
            {
                ans[i]=temp-20;
                jin[i+1]=1;
            }
            else
                ans[i]=temp;
        }
        if(ans[len])
            cout<=0;i--)
            cout<