摘要: /********************LightOJ 1050Author:Cdegree********************/#include #include #include #include #include #include #include #include #include #include #include #include #include #define X first#define Y second#define sqr(x) (x)*(x)#pragma comment(linker,"/STACK:102400000,102400000") 阅读全文
posted @ 2013-10-02 15:03 degree 阅读(172) 评论(0) 推荐(0)
摘要: 其实就是求最长回文序列的长度m,然后答案=n-m;/********************LightOJ 1033Author:Cdegree********************/#include #include #include #include #include #include #include #include #include #include #include #include #include #define X first#define Y second#define sqr(x) (x)*(x)#pragma comment(linker,"/STACK:1 阅读全文
posted @ 2013-10-01 20:07 degree 阅读(235) 评论(0) 推荐(0)
摘要: /********************LightOJ 1032Author:Cdegree********************/#include #include #include #include #include #include #include #include #include #include #include #include #include #define X first#define Y second#define sqr(x) (x)*(x)#pragma comment(linker,"/STACK:102400000,102400000") 阅读全文
posted @ 2013-10-01 20:00 degree 阅读(218) 评论(0) 推荐(0)
摘要: /********************LightOJ 1031Author:Cdegree********************/#include #include #include #include #include #include #include #include #include #include #include #include #include #define X first#define Y second#define sqr(x) (x)*(x)#pragma comment(linker,"/STACK:102400000,102400000") 阅读全文
posted @ 2013-10-01 16:00 degree 阅读(295) 评论(0) 推荐(0)
摘要: /********************Author:Cdegree********************/#include #include #include #include #include #include #include #include #include #include #include #include #include #define X first#define Y second#define sqr(x) (x)*(x)#pragma comment(linker,"/STACK:102400000,102400000")using namesp 阅读全文
posted @ 2013-10-01 15:27 degree 阅读(165) 评论(0) 推荐(0)
摘要: 平均访问时间为sum(t)/n,期望次数为n/n1#include #include #include #include #include #include #include #include #include #include #include #include #include #define X first#define Y second#define sqr(x) (x)*(x)#pragma comment(linker,"/STACK:102400000,102400000")using namespace std;const double PI = acos( 阅读全文
posted @ 2013-10-01 15:26 degree 阅读(169) 评论(0) 推荐(0)
摘要: #include #include #include #include #include #include #include #include #include #include #include #include #include #define X first#define Y second#define sqr(x) (x)*(x)#pragma comment(linker,"/STACK:102400000,102400000")using namespace std;const double PI = acos(-1.0);map::iterator it;ty 阅读全文
posted @ 2013-10-01 15:04 degree 阅读(188) 评论(0) 推荐(0)
摘要: #include #include #include #include #include #include #include #include #include #include #include #include #include #define X first#define Y second#define sqr(x) (x)*(x)#pragma comment(linker,"/STACK:102400000,102400000")using namespace std;const double PI = acos(-1.0);map::iterator it;ty 阅读全文
posted @ 2013-10-01 15:04 degree 阅读(152) 评论(0) 推荐(0)
摘要: #include #include #include #include #include #include #include #include #include #include #include #include #include #define X first#define Y second#define sqr(x) (x)*(x)#pragma comment(linker,"/STACK:102400000,102400000")using namespace std;const double PI = acos(-1.0);map::iterator it;ty 阅读全文
posted @ 2013-10-01 15:03 degree 阅读(153) 评论(0) 推荐(0)
摘要: #include #include #include #include #include #include #include #include #include #include #include #include #include #define X first#define Y second#define sqr(x) (x)*(x)#pragma comment(linker,"/STACK:102400000,102400000")using namespace std;const double PI = acos(-1.0);map::iterator it;ty 阅读全文
posted @ 2013-10-01 15:02 degree 阅读(131) 评论(0) 推荐(0)