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 <…

 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: 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…

 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…

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…

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…

[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…

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 bitwise operation). Operation 2: OR opn L R H…

题目大意:给n个0~15之间的数,有3种更新操作,1种询问操作.3种更新操作是:1.让某个闭区间的所有数字与一个0~15之间的数字进行逻辑与运算:2.让某个闭区间的所有数字与一个0~15之间的数字进行逻辑或运算:3.让某个闭区间的所有数字与一个0~15之间的数字进行异或运算.一种询问操作是询问某个闭区间的所有数字之和. 题目分析:所有的输入数字都是在0~15之间,可以二进制中的每一位建立线段树,建立四棵.3种更新操作实际上只有两种,即区间置位和区间反转.用两个懒惰标记,当进行区间置位更新的时候,…

题意: 给出区间与.或.异或\(x\)操作,还有询问区间和. 思路: 因为数比较小,我们给每一位建线段树,这样每次只要更新对应位的答案. 与\(0\)和或\(1\)相当于重置区间,异或\(1\)相当于翻转区间,那么设出两个\(lazy\)搞一下.注意父区间\(pushdown\)重置标记时,子区间的翻转标记要清空. 代码: #include <map> #include <set> #include <queue> #include <cmath> #inc…