Problem 2105 Digits Count Accept: 302 Submit: 1477 Time Limit: 10000 mSec Memory Limit : 262144 KB Problem Description Given N integers A={A[0],A[1],-,A[N-1]}. Here we have some operations: Operation 1: AND opn L R Here opn, L and R are integers. For…

http://acm.fzu.edu.cn/problem.php?pid=2105 Problem Description Given N integers A={A[0],A[1],...,A[N-1]}. Here we have some operations: Operation 1: AND opn L R Here opn, L and R are integers. For L≤i≤R, we do A[i]=A[i] AND opn (here "AND" is bi…

 FZU 2105  Digits Count Time Limit:10000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Practice Description Given N integers A={A[0],A[1],...,A[N-1]}. Here we have some operations: Operation 1: AND opn L R Here opn, L and R are intege…

 Problem 2105 Digits Count Accept: 444    Submit: 2139 Time Limit: 10000 mSec    Memory Limit : 262144 KB  Problem Description Given N integers A={A[0],A[1],...,A[N-1]}. Here we have some operations: Operation 1: AND opn L R Here opn, L and R are int…

[题目链接] http://acm.fzu.edu.cn/problem.php?pid=2105 [题目大意] 给出一个序列,数字均小于16,为正数,每次区间操作可以使得 1. [l,r]区间and一个数 2. [l,r]区间or一个数 3. [l,r]区间xor一个数 4. [l,r]区间查询和 操作数均为小于16的非负整数 [题解] 由于操作数很小,因此我们可以按位维护四棵线段树,表示二进制中的第i位, 对于and操作,只有当and的当前位为0时才对区间有影响,效果是将区间全部变为0, 对…

Description 题目描述 Given N integers A={A[0],A[1],...,A[N-1]}. Here we have some operations: Operation 1: AND opn L R Here opn, L and R are integers. For L≤i≤R, we do A[i]=A[i] AND opn (here "AND" is bitwise operation). Operation 2: OR opn L R Here…

Subsequence Count Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 256000/256000 K (Java/Others) Problem Description Given a binary string S[1,...,N] (i.e. a sequence of 0's and 1's), and Q queries on the string. There are two types of querie…

题意:对一串数字进行抑或某数,和某数,或某数,统计某区间和的操作. 思路:因为化成二进制就4位可以建4颗线段树,每颗代表一位二进制. and 如果该为是1  直接无视,是0则成段赋值为0: or  如果是0 无视,是1则成段赋值为1: xor 成段亦或,1个数和0个数交换: sum 求和: #include<cstdio> #include<cstring> #include<algorithm> #include<vector> #include <…

DP式很容易得到,发现是线性递推形式,于是可以矩阵加速.又由于是区间形式,所以用线段树维护. https://www.cnblogs.com/Miracevin/p/9124511.html 关键在于证明区间操作中,可以直接在打标记的位置翻转矩阵两行两列. 上面网址用代数形式证了一遍,这里考虑从矩阵本身解释. 由线代内容可知,将一个矩阵作初等行变换,相当于将其左乘一个作了相应初等列变换的单位矩阵.同理将一个矩阵作初等列变换,相当于将其又乘一个作了相应初等行变换的单位矩阵. 这里,左乘的矩阵$T=…

Description Given N integers A={A[0],A[1],...,A[N-1]}. Here we have some operations: Operation 1: AND opn L R Here opn, L and R are integers. For L≤i≤R, we do A[i]=A[i] AND opn (here "AND" is bitwise operation). Operation 2: OR opn L R Here opn,…