摘要: 素数判定Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 10 Accepted Submission(s) : 4Font: Times New Roman | Verdana | GeorgiaFont Size: ← →Problem Description对于表达式n^2+n+41,当n在(x,y)范围内取整数值时(包括x,y)(-39using namespace std;int shu(int a)//判断素数{ int i=2... 阅读全文
posted @ 2013-07-17 21:30 一只蚊子 阅读(260) 评论(1) 推荐(0)
摘要: 最小公倍数Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 11 Accepted Submission(s) : 2Font: Times New Roman | Verdana | GeorgiaFont Size: ← →Problem Description给定两个正整数,计算这两个数的最小公倍数。Input输入包含多组测试数据,每组只有一行,包括两个不大于1000的正整数.Output对于每个测试用例,给出这两个数的最小公倍数,每个实例 阅读全文
posted @ 2013-07-13 22:09 一只蚊子 阅读(168) 评论(0) 推荐(0)
摘要: a/b + c/dTime Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 0 Accepted Submission(s) : 0Font: Times New Roman | Verdana | GeorgiaFont Size: ← →Problem Description给你2个分数,求他们的和,并要求和为最简形式。 Input输入首先包含一个正整数T(T using namespace std;int main() {int n; cin> 阅读全文
posted @ 2013-07-13 21:11 一只蚊子 阅读(358) 评论(0) 推荐(0)
摘要: Problem DescriptionI have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B. InputThe first line of the input contains an integer T(1#include using namespace std;int main(){ void sum(string &a,string &b,string &c);int n; cin>>n;//输 阅读全文
posted @ 2013-07-13 16:36 一只蚊子 阅读(126) 评论(0) 推荐(0)