08 2013 档案

摘要:High School/Jr.High 10 PRINT "HELLO WORLD" 20 ENDFirst year in College program Hello(input, output) begin writeln('Hello World') end.Senior year in College (defun hello (print (cons 'Hello (list 'World))))New professional #include void main(void) { char *message[] = {" 阅读全文
posted @ 2013-08-30 18:01 天天AC 阅读(193) 评论(0) 推荐(0) 编辑
摘要:void catalan() //求卡特兰数{ int i, j, len, carry, temp; a[1][0] = b[1] = 1; len = 1; for(i = 2; i = 0; j--) //除法 { temp = carry*10 + a[i][j]; a[i][j] = temp/(i+1); carry = temp%(i+1); } while(!a[i][len-1]) //高位零处理 len --; b[i] = len; }} 阅读全文
posted @ 2013-08-28 20:43 天天AC 阅读(371) 评论(0) 推荐(0) 编辑
摘要:PillsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 746 Accepted Submission(s): 490Problem DescriptionAunt Lizzie takes half a pill of a certain medicine every day. She starts with a bottle that contains N pills.On the first day, she removes a... 阅读全文
posted @ 2013-08-28 20:39 天天AC 阅读(261) 评论(0) 推荐(0) 编辑
摘要:ScoreTime Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 778 Accepted Submission(s): 340Problem Description大家都知道,pfz是“成电杰出学生”,在成电杰出学生的颁奖典礼上,lxh和pfz都没有听台上在说什么,而是在下面讨论当晚的美式足球比赛,lxh预测说纽约巨人队今晚将会得到11分,pfz马上说不可能。因为通常来说美式足球比赛的得分只有3分和7分两种形式,无论怎么得分都不可能得到11分... 阅读全文
posted @ 2013-08-28 20:34 天天AC 阅读(181) 评论(0) 推荐(0) 编辑
摘要:/* C,翻译自 C++ 代码 */# include # includeusing namespace std;const int maxCount = 100;//程序允许的最多珠子数//FILE *fin, *fout;//输入输出文件指针int n;//珠子数int v[maxCount * 2];//v[i]表示第(i % n)颗珠子的能量,%表示求余int energy[maxCount * 2][maxCount * 2];//energy[i][j]表示从第i颗珠子到第j颗珠子组成的珠子片断能够得到的最大能量//初始化void init(){ //fopen ( "e 阅读全文
posted @ 2013-08-10 16:04 天天AC 阅读(315) 评论(0) 推荐(0) 编辑
摘要:Time Limit: 1000 MSMemory Limit: 65536 KTotal Submit: 70(43 users)Total Accepted: 50(40 users)Rating:Special Judge: NoDescription从m个不同元素中取出n (n ≤ m)个元素的所有组合的个数,叫做从m个不同元素中取出n个元素的组合数。组合数的计算公式如下:C(m, n) = m!/((m - n)!n!)现在请问,如果将组合数C(m, n)写成二进制数,请问转这个二进制数末尾有多少个零。Input第一行是测试样例的个数T,接下来是T个测试样例,每个测试样例占一行,有两 阅读全文
posted @ 2013-08-10 15:56 天天AC 阅读(343) 评论(0) 推荐(0) 编辑
摘要:#include #include #include #include #include #include #define RE(x) (x)^1#define INF 0x3fffffff#define MAXN 205using namespace std;int N, M, G[MAXN][MAXN], dis[MAXN*MAXN];int dir[2][2] = {0, 1, 1, 0};int idx, head[MAXN*MAXN];const int sink = MAXN*MAXN-1, source = MAXN*MAXN-2;struct Edge{ int v, c... 阅读全文
posted @ 2013-08-02 19:49 天天AC 阅读(161) 评论(0) 推荐(0) 编辑

点击右上角即可分享
微信分享提示