西山居初赛第四场1001

本文主要是介绍西山居初赛第四场1001，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

叛逆的小明

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 81 Accepted Submission(s): 61

Problem Description

叛逆期的小明什么都喜欢反着做，连看数字也是如此（负号除外），比如：
小明会把1234它看成4321；把-1234看成-4321；把230看成032 (032=32)；把-230看成-032(-032=-32)。

现在，小明做了一些a+b和a-b的题目(a, b为整数且不含前导0)，如果给你这些题目正确的答案，你能猜出小明会做得到什么答案吗？

Input

输入第一行为一个正整数T(T<=10000)，表示小明共做了T道题。
接下来T行，每行是两个整数x,y(-1000000<=x, y<=1000000), x表示a+b的正确答案，y表示a-b的正确答案。
输入保证合法，且不需考虑a或b是小数的情况。

Output

输出共T行，每行输出两个整数s t，之间用一个空格分开，其中s表示小明将得到的a+b答案，t表示小明将得到的a-b答案。

Sample Input

Sample Output

  38 24
7 7
-19 -23

#include<stdio.h>
long huiwen1(long n)
{long sum,i=0;if(n<10) return n;while(n>9){if(i==0)sum=n%10;n=n/10;sum=n%10+sum*10;i++;}return sum;
}
long huiwen2(long n)
{long sum=0,i=0;if(n>-10)  return n;while(n<-9){if(i==0)sum=n%10;n=n/10;sum=n%10+sum*10;i++;}return sum;
}
main()
{long T,a,b,x,y;scanf("%ld",&T);while(T--){scanf("%ld%ld",&x,&y);b=(x-y)/2;a=x-b;if((a>=0)&&(b>=0))printf("%ld %ld\n",huiwen1(a)+huiwen1(b),huiwen1(a)-huiwen1(b));if(a<0&&b<0)printf("%ld %ld\n",huiwen2(a)+huiwen2(b),huiwen2(a)-huiwen2(b));if(a>=0&&b<0)printf("%ld %ld\n",huiwen1(a)+huiwen2(b),huiwen1(a)-huiwen2(b));if(a<0&&b>=0)printf("%ld %ld\n",huiwen2(a)+huiwen1(b),huiwen2(a)-huiwen1(b));}return 0;
}
#include<stdio.h>
long huiwen1(long n)
{long sum,i=0;if(n<10) return n;while(n>9){if(i==0)sum=n%10;n=n/10;sum=n%10+sum*10;i++;}return sum;
}
long huiwen2(long n)
{long sum=0,i=0;if(n>-10)  return n;while(n<-9){if(i==0)sum=n%10;n=n/10;sum=n%10+sum*10;i++;}return sum;
}
main()
{long T,a,b,x,y;scanf("%ld",&T);while(T--){scanf("%ld%ld",&x,&y);b=(x-y)/2;a=x-b;if((a>=0)&&(b>=0))printf("%ld %ld\n",huiwen1(a)+huiwen1(b),huiwen1(a)-huiwen1(b));if(a<0&&b<0)printf("%ld %ld\n",huiwen2(a)+huiwen2(b),huiwen2(a)-huiwen2(b));if(a>=0&&b<0)printf("%ld %ld\n",huiwen1(a)+huiwen2(b),huiwen1(a)-huiwen2(b));if(a<0&&b>=0)printf("%ld %ld\n",huiwen2(a)+huiwen1(b),huiwen2(a)-huiwen1(b));}return 0;
}