hdu 2095 find your present (2)
2010-10-18 22:20
323 查看
http://acm.hdu.edu.cn/showproblem.php?pid=2095
方法很多:(1)查找,数组设为499999超时了!ac不了!
(2)you can assume that only one number appear odd times.根据这句话可以想想更高效的算法,思考中……
(3)用异或运算符
异或运算符的用法:
整数的异或是先把它们化成二进制,再按位异或。比如3^5, 3=011,5=101,两数按位异或后为
110,即6。
几个数异或满足交换律。2^3^2=2^2^3=0^3=3.
两个相同的数异或为0,普通数都出现了偶数次,所以它们异或后都是0,而0与那个特别数异或后还是那个特殊数。
Accepted 2095 562MS 180K 226 B G++ rll
代码
方法很多:(1)查找,数组设为499999超时了!ac不了!
(2)you can assume that only one number appear odd times.根据这句话可以想想更高效的算法,思考中……
(3)用异或运算符
异或运算符的用法:
整数的异或是先把它们化成二进制,再按位异或。比如3^5, 3=011,5=101,两数按位异或后为
110,即6。
几个数异或满足交换律。2^3^2=2^2^3=0^3=3.
两个相同的数异或为0,普通数都出现了偶数次,所以它们异或后都是0,而0与那个特别数异或后还是那个特殊数。
Accepted 2095 562MS 180K 226 B G++ rll
代码
#include<stdio.h> //异或运算的运用 int main() { int t,a,sum; while(scanf("%d",&t)&&t!=0) { scanf("%d",&sum); t--; while(t--) { scanf("%d",&a); sum^=a; } printf("%d\n",sum); } return 0; }
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