zoj 3432 与运算运用

  1. /****************************************************
  2.  该题充分运用与运算的特点，成对消除，单的最后留下
  3.  ****************************************************/
  5. #include <stdio.h>
  6. #include <string.h>
  7. #include <algorithm>
  8. using namespace std;
  10. int main()
  11. {
  12. 	char str1[10], str2[10];
  13. 	int n;
  14. 	while(~scanf("%d", &n) )
  15. 	{
  16. 		getchar();
  18. 		gets(str1);
  19. 		for(int i=0; i<2*n-2; i++)
  20. 		{
  21. 			gets(str2);
  22. 			for(int j=0; j<7; j++)
  23. 			str1[j] = str1[j] ^ str2[j];				//两个字符串各自与运算，可以判定最后剩下的不成对的一个
  24. 		}
  25. 		printf("%s
", str1);
  26. 	}
  27. }
  29. /*
  30. Description
  31. Alice bought a lot of pairs of socks yesterday. But when she went home, she found that she has lost one of them. Each sock has a name which contains exactly 7 charaters.
  34. Alice wants to know which sock she has lost. Maybe you can help her.
  36. Input
  37. There are multiple cases. The first line containing an integer n (1 <= n <= 1000000) indicates that Alice bought n pairs of socks. For the following 2*n-1 lines, each line is a string with 7 charaters indicating the name of the socks that Alice took back.
  39. Output
  40. The name of the lost sock.
  42. Sample Input
  43. 2
  44. aabcdef
  45. bzyxwvu
  46. bzyxwvu
  47. 4
  48. aqwerty
  49. eas fgh
  50. aqwerty
  51. easdfgh
  52. easdfgh
  53. aqwerty
  54. aqwerty
  55. 2
  56. 0x0abcd
  57. 0ABCDEF
  58. 0x0abcd
  59. Sample Output
  60. aabcdef
  61. eas fgh
  62. 0ABCDEF
  63. */





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    原文地址：https://www.cnblogs.com/sober-reflection/p/cd588e8df002561ac9f5a9bfed72c57f.html