Description It is well-known that for any n there are exactly four n-digit numbers (including ones with leading zeros) that are self-squares: the last ndigits of the square of such number are equal to the number itself. These four numbers are always…
做了三天,,,终于a了... 11724203 2014-09-25 09:37:44 Accepted 5033 781MS 7400K 4751 B G++ czy Building Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 1257 Accepted Submission(s): 358 Special Judg…
题意:http://acdream.info/problem?pid=1112 Problem Description Here is Alice and Bob again ! Alice and Bob are playing a game. There are several numbers.First, Alice choose a number n.Then he can replace n (n > 1)with one of its positive factor but not…
Alice and Bob Time Limit: 1000ms Memory limit: 65536K 有疑问?点这里^_^ 题目描述 Alice and Bob like playing games very much.Today, they introduce a new game. There is a polynomial like this: (a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1). T…
Alice and Bob Time Limit: 1000ms Memory limit: 65536K 题目描述 Alice and Bob like playing games very much.Today, they introduce a new game. There is a polynomial like this: (a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1). Then Alice as…
http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2608 Alice and Bob Time Limit: 1000ms Memory limit: 65536K 有疑问?点这里^_^ 题目描述 Alice and Bob like playing games very much.Today, they introduce a new game. There is a polynomial li…
Alice and Bob Time Limit: 1000ms Memory limit: 65536K 题目描述 Alice and Bob like playing games very much.Today, they introduce a new game. There is a polynomial like this: (a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1). Then Alice…
Alice and Bob Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2901 Accepted Submission(s): 941 Problem Description Alice and Bob's game never ends. Today, they introduce a new game. In this…
Minimum palindrome Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 260 Accepted Submission(s): 127 Problem Description Setting password is very important, especially when you have so many "in…
又见Alice and Bob 时间限制:1000 ms | 内存限制:65535 KB 难度:3 描述 集训生活如此乏味,于是Alice和Bob发明了一个新游戏.规则如下:首先,他们得到一个集合包含n个特定的整数,接着他们轮流做以下操作,每一次操作,Alice或者Bob(轮到谁就是谁)会从集合中选择两个整数x 和 y ,(但是集合中不能包含| x - y|),接着他就会把整数|x - y| 加入集合,因此,集合中的数据多加了一个…… 如果当前玩家不能执行操作了,他就输了.问题是如果Al…
Alice and Bob Accepted : 133 Submit : 268 Time Limit : 1000 MS Memory Limit : 65536 KB Problem Description The famous "Alice and Bob" are playing a game again. So now comes the new problem which need a person smart as you to decide the winn…
题目描述 Alice and Bob like playing games very much.Today, they introduce a new game. There is a polynomial like this: (a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1). Then Alice ask Bob Q questions. In the expansion of the Polynomial, G…
Alice and Bob Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 147 Accepted Submission(s): 22 Problem Description As you know, Alice and Bob always play game together, and today they get a…
Alice and Bob Time Limit : 10000/5000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 5 Accepted Submission(s) : 1 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Alice and Bob's game nev…
Alice and Bob Time Limit:3000MS Memory Limit:128000KB 64bit IO Format:%lld & %llu Submit Status Practice ACdream 1112 Description Here is Alice and Bob again ! Alice and Bob are playing a game. There are several numbers. First, Alice choose…
Alice and Bob Time Limit: 1000ms Memory limit: 65536K 有疑问?点这里^_^ 题目描述 Alice and Bob like playing games very much.Today, they introduce a new game. There is a polynomial like this: (a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1). T…
Alice and Bob Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3716 Accepted Submission(s): 1179 Problem Description Alice and Bob's game never ends. Today, they introduce a new game. In this game…
题目链接:1484 - Alice and Bob's Trip 题意:BOB和ALICE这对狗男女在一颗树上走,BOB先走,BOB要尽量使得总路径权和大,ALICE要小,可是有个条件,就是路径权值总和必须在[L,R]之间,求终于这条路径的权值. 思路:树形dp,dp[u]表示在u结点的权值,往下dfs的时候顺带记录下到根节点的权值总和,然后假设dp[v] + w + sum 在[l,r]内,就是能够的,状态转移方程为 dp[u] = max{dp[v] + w }(bob) dp[u] = m…
