2012年11月1日
交换a,b
摘要： /*交换两个数不用中间变量*/#include<iostream>using namespace std;int main(){ int a,b; scanf("%d%d",&a,&b); a=a+b; b=a-b; a=a-b; printf("%d %d\n",a,b); return 0;} 阅读全文
posted @ 2012-11-01 23:34 myth_HG 阅读(121) 评论(0) 推荐(0) 编辑
N^N hdu1060
摘要： Leftmost DigitTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8784 Accepted Submission(s): 3391Problem DescriptionGiven a positive integer N, you should output the leftmost digit of N^N. InputThe input contains several test cases. The first ... 阅读全文
posted @ 2012-11-01 23:23 myth_HG 阅读(649) 评论(2) 推荐(0) 编辑