浙大计算机研究生保研复试上机考试-2011年  贪心: 注意:输入输出用scanf  printf 可以加快速度,用cin WA #include<iostream> #include<cstring> #include<cstdio> #include<string> #include<cmath> #include<algorithm> using namespace std; #define MAX 1000000 __int6…

1037. Magic Coupon (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue The magic shop in Mars is offering some magic coupons. Each coupon has an integer N printed on it, meaning that when you use this coupon with a product, you m…

1037 Magic Coupon (25 分) The magic shop in Mars is offering some magic coupons. Each coupon has an integer N printed on it, meaning that when you use this coupon with a product, you may get N times the value of that product back! What is more, the sh…

1037 Magic Coupon(25 分) The magic shop in Mars is offering some magic coupons. Each coupon has an integer N printed on it, meaning that when you use this coupon with a product, you may get N times the value of that product back! What is more, the sho…

hdu 1153 magic bitstrings 题目大意 一个质数p,现在让你求一个p-1长度的“01魔法串”.关于这个魔法串是这么定义的:     我们现在把这个串经过一段处理变成一个长宽均为p-1的矩阵,对于第i行的串,是由原来的串按每i位取得的.如果这个矩阵每行的串满足:和原来的串相等或是原来的串按位取反,我们就称这个串是魔法串.(说了一大堆如果还没看懂就去问题底下的Discuss看吧….) 题解 一开始热衷于讨论第一行和最后一行的关系…结果什么也没看出来…后来看了别人的题解,才发现自…

1037 Magic Coupon (25 分)   The magic shop in Mars is offering some magic coupons. Each coupon has an integer N printed on it, meaning that when you use this coupon with a product, you may get N times the value of that product back! What is more, the…

Source: PAT A1037 Magic Coupon (25 分) Description: The magic shop in Mars is offering some magic coupons. Each coupon has an integer N printed on it, meaning that when you use this coupon with a product, you may get N times the value of that product…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2274 Magic WisKey Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 568    Accepted Submission(s): 323 Problem Description On New Year Festival, Liu…

http://acm.hdu.edu.cn/showproblem.php?pid=4605 可以离线求解 把所以可能出现的 magic ball  放在一个数组里(去重),从小到大排列 先不考虑特殊情况,对二叉树进行dfs 搜索的过程中需要维护各个magic ball到当前节点的概率 维护:根据当前节点大小 和要去左子树还是右子树的情况,可以得到magic数组中哪个段的x和y需要同时加上多少 可以用线段树维护 特殊情况: 1,根节点一定可以到达 2,到达不了的情况需要标记 代码: #inclu…

主题链接: http://acm.hdu.edu.cn/showproblem.php?pid=4016 Magic Bitwise And Operation Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others) Total Submission(s): 1315    Accepted Submission(s): 504 Problem Description Given n…