《ACM程序设计》书中题目―N

Description

The Antique Comedians of Malidinesia prefer comedies to tragedies. Unfortunately, most of the ancient plays are tragedies. Therefore the dramatic advisor of ACM has decided to transfigure some tragedies into comedies. Obviously, this work is very hard because the basic sense of the play must be kept intact, although all the things change to their opposites. For example the numbers: if any number appears in the tragedy, it must be converted to its reversed form before being accepted into the comedy play.

Reversed number is a number written in arabic numerals but the order of digits is reversed. The first digit becomes last and vice versa. For example, if the main hero had 1245 strawberries in the tragedy, he has 5421 of them now. Note that all the leading zeros are omitted. That means if the number ends with a zero, the zero is lost by reversing (e.g. 1200 gives 21). Also note that the reversed number never has any trailing zeros.

ACM needs to calculate with reversed numbers. Your task is to add two reversed numbers and output their reversed sum. Of course, the result is not unique because any particular number is a reversed form of several numbers (e.g. 21 could be 12, 120 or 1200 before reversing). Thus we must assume that no zeros were lost by reversing (e.g. assume that the original number was 12).

Input

The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case consists of exactly one line with two positive integers separated by space. These are the reversed numbers you are to add.

Output

For each case, print exactly one line containing only one integer - the reversed sum of two reversed numbers. Omit any leading zeros in the output.

Sample Input

3

24 1

4358 754

305 794

Sample Output

34

1998

1

1、题意：输入两个数，让你把他们倒置然后相加再倒置输出。例如输入123与456 即是变为321+654等于975然后输出579。

2、思路：只要解决了数的倒置本题就解决了，利用％10取余然后数每次除以10循环，然后取出来的数每次乘以10依次循环直到取完数即可得到倒置的数。

3、代码：

#include<iostream>  using namespace std;int main()   {       int n,t,t2,x,x2,y,y2;    cin>>n;       while(n--)      {          cin>>x>>y;x2=0;y2=0;t2=0;while(x)      {           x2=x2*10+x%10;           x=x/10;       }while(y)      {           y2=y2*10+y%10;           y=y/10;       }           t=x2+y2;        while(t)      {           t2=t2*10+t%10;           t=t/10;       }cout<<t2<<endl;      }    return 0;   }

4、总结：其实将倒置的过程写成一个函数，然后每次需要的气候调用即可，结果每次都写一次导致代码过长，下次遇到类似的题一定记住写成函数然后调用，使得代码简洁美观。