1205_Martian Addition
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
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#include<iostream>
#include<vector>
#include<string>
using namespace std;
void equ(string,string,string,vector<int>);
int main()
{
string s0("0123456789abcdefghij");
string s1,s2;
vector<int> v1;
int num;
while(cin>>s1>>s2)
{
if(s1.size()==s2.size())//两个数长度相等
equ(s0,s1,s2,v1);
else //两数长度不相等（题中并没有举出两数不相等的例子，但是只有这样才能ac...）
{
if(s1.size()>s2.size())
{
num=s1.size()-s2.size();
for(int i=0;i!=num;i++)
s2.insert( 0,"0");//在短的前面补"0"，和长的size相同（技巧！！！）
}
else
{
num=s2.size()-s1.size();
for(int i=0;i!=num;i++)
s1.insert( 0,"0");
}
equ(s0,s1,s2,v1);
}
}
return 0;
}
void equ(string s0,string s1,string s2,vector<int> v1)
{
int a,b,c,d;
c=0;
for(int i=s1.size()-1;i>=0;i--)
{
if(s1[i]<=57)//将string转换为实际值
a=s1[i]-48;
else
a=s1[i]-87;
if(s2[i]<=57)
b=s2[i]-48;
else
b=s2[i]-87;
d=a+b+c;
if(d>=20)//进位，因为是两个数相加，所以进位最多为1；
{
d-=20;
c=1;
}
else
c=0;
v1.push_back(d);
}
if(c==1)
v1.push_back(c);//最高为进位
for(vector<int>::iterator iter=v1.end()-1;iter>v1.begin();--iter)//倒序输出时，若是iter>=v1.begin()；输出最后一个后，会地址越界，只能这样
cout<<s0[*iter];
cout<<s0[v1[0]]<<endl;
v1.clear();
}