MZL's xor Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 249    Accepted Submission(s): 187 Problem Description MZL loves xor very much.Now he gets an array A.The length of A is n.He wants to k…

Problem Description   MZL loves xor very much.Now he gets an array A.The length of A ≤i,j≤n) The xor of an array B is defined as B1 xor B2...xor Bn Input Multiple test cases, the first line contains an integer T(no more than ), indicating the number…

MZL's xor Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 911    Accepted Submission(s): 589 Problem Description MZL loves xor very much.Now he gets an array A.The length of A is n.He wants to k…

题意:给一个序列A,设序列B的中的元素有(Ai+Aj)(1≤i,j≤n),那么求B中所有元素的异或之和.而序列A是这样来的:A1=0,Ai=(Ai−1∗m+z) mod l. 思路:相同的元素异或结果为0,所以可以去掉,也就是剩下A中的元素ai+ai那些而已. 小心乘法会溢出 #include <bits/stdc++.h> #define INF 0x7f7f7f7f #define pii pair<int,int> #define LL long long using nam…

MZL's xor Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 800    Accepted Submission(s): 518 Problem Description MZL loves xor very much.Now he gets an array A.The length of A is n.He wants to…

题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=5344 #include<stdio.h> #include<cstring> ; int A[MAXN]; void bas( int n, int m, int z, int l){ ;i < n; ++i){ A[i] = ( (]*m + z )% l; } } void Cal(int n){ ; ;i < n; ++i){ sum = sum ^ (A[i]*);…

[HDU 5351 MZL's Border]题意 定义字符串$f_1=b,f_2=a,f_i=f_{i-1}f_{i-2}$. 对$f_n$的长度为$m$的前缀$s$, 求最大的$k$满足$s[1]=s[m-k+1],s[2]=s[m-k+2]...s[k]=s[m]$ Solution 先列出前几个字符串 $f_3=ab,f_4=aba,f_5=abaab,f_6=abaababa,f_7=abaababaabaab$. 对于n,m. 假设$f_{n-1}$的长度大于等于$m$,那么相当于求…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud MZL's xor Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 488    Accepted Submission(s): 342 Problem Description MZL loves xor very muc…

MZL's xor Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/65536K (Java/Other) Total Submission(s) : 1   Accepted Submission(s) : 1 Problem Description MZL loves xor very much.Now he gets an array A.The length of A is n.He wants to know t…

题目链接: Hdu 5352 MZL's City 题目描述: 有n各节点,m个操作.刚开始的时候节点都是相互独立的,一共有三种操作: 1:把所有和x在一个连通块内的未重建过的点全部重建. 2:建立一条双向路(x,y) 3:又发生了地震,p条路被毁. 问最后最多有多少个节点被重建,输出重建节点的最小字典序. 解题思路: 这几天正好在学匹配,但是昨天下午还是没有看出来这个是匹配题目.看了题解扪心自问了自己三次,是不是傻.就是把每个需要重建的节点x拆成k个点,然后对每个拆分后的点和与拆点在同一连通块…