Xor Sum(HDU4825 + 字典树) 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4825 题目: 题意: 先给你n个数,再进行q次查询,每次查询数s与原来给的n个数异或和最大的数。 思路: 建一棵字典树,将n个数转换成二进制存进去,每次查询时,对s的每一位进行匹配(从高位开始匹配,毕竟你低位再大也没...
root=tir[root][id]; }// printf("\n");LL sum2=0; LL sum1=1;for(LL i=0;i<=31;i++){// printf("*%d ",num[i]);sum2+=num[i]*sum1; sum1*=2; }//printf("\n");returnsum2; }voidallbegin(){for(inti=0;i<=top+10;i++){for(intj=0;j<=2;j++){ tir[i][j]=0;...
This paper is an attempt to propose an improved version of a famous feature descriptor, called Xor-sum code, to obtain an enhanced recognition accuracy. The proposed approach incorporates the curvature information into the conventional Gabor filter, to facilitate discriminative iris feature representation...
Problem C: 异或和(xorsum) 上一题正确率:25.20%下一题 " Time Limit1秒/Second(s)Memory Limit128兆字节/Megabyte(s) 提交总数127正确数量32 裁判形式标准裁判/Standard Judge我的状态尚未尝试 难度分类标签递推 Description 小可可在五年级暑假开始学习编程,编程语言中有一种“按位异或(xor)”的运算引...
HDU 4825 Xor Sum Xor Sum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 3633 Accepted Submission(s): 1590 Probl c语言 题型 hdu4825---Xor Sum(Trie + 贪心) Problem Description Zeus 和 Prometheus 做了一个游戏,Prometheus 给 Zeus 一...
// Problem: D - Xor Sum 2// URL: https://atcoder.jp/contests/abc098/tasks/arc098_b// Author: Pengfei Xu#include<bits/stdc++.h>using namespace std;using ll = long long;const int maxn=2e5+5;int a[maxn];int solve(){ int n; cin >> n; for...
2014年百度之星程序设计大赛 - 资格赛 第三题 Xor Sum,小记:蛋呢,取longlong的低30,32,34位都WA,取31位才AC。。。思路:根据求数组中两个数异或最大值,参考代码:#include<stdio.h>#include<string.h>#defineMAX(a,b)((a)>(b)?(a):(b))#defineNODE32000
Sum of XOR Functions 类似的形式还有 \sum_{i=1}^{n}\sum_{j=i}^n\bigoplus_{k=i}^{j}a_k(j-i+1)\\ 即转化为 \sum_{i=1}^{n}\sum_{j=i}^nb_j\oplus b_{i-1}(j-(i-1))\\ 只需再维护一个权值数组 w[2] 即可 代码 #include<bits/stdc++.h> using namespace std;...
Prometheus 为了让 Zeus 看到人类的伟大,随即同意 Zeus 可以向人类求助。你能证明人类的智慧么? Input...
# 题解_Atcoder_Xor Sum 2 ## 1:题意 ### Problem Statement There is an integer sequence A of length N. Find the number of the pairs of integers *l* and r ($1 \leq l \leq r \leq N$) that satisfy the following condition: ...