zoukankan      html  css  js  c++  java
  • CodeForces 242E二维线段树

                                                                                           E. XOR on Segment

    You've got an array a, consisting of n integers a1, a2, ..., an. You are allowed to perform two operations on this array:

    1. Calculate the sum of current array elements on the segment [l, r], that is, count value al + al + 1 + ... + ar.
    2. Apply the xor operation with a given number x to each array element on the segment [l, r], that is, execute . This operation changes exactly r - l + 1 array elements.

    Expression means applying bitwise xor operation to numbers x and y. The given operation exists in all modern programming languages, for example in language C++ and Java it is marked as "^", in Pascal — as "xor".

    You've got a list of m operations of the indicated type. Your task is to perform all given operations, for each sum query you should print the result you get.

    Input

    The first line contains integer n (1 ≤ n ≤ 105) — the size of the array. The second line contains space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 106) — the original array.

    The third line contains integer m (1 ≤ m ≤ 5·104) — the number of operations with the array. The i-th of the following m lines first contains an integer ti (1 ≤ ti ≤ 2) — the type of the i-th query. If ti = 1, then this is the query of the sum, if ti = 2, then this is the query to change array elements. If the i-th operation is of type 1, then next follow two integers li, ri (1 ≤ li ≤ ri ≤ n). If the i-th operation is of type 2, then next follow three integers li, ri, xi (1 ≤ li ≤ ri ≤ n, 1 ≤ xi ≤ 106). The numbers on the lines are separated by single spaces.

    Output

    For each query of type 1 print in a single line the sum of numbers on the given segment. Print the answers to the queries in the order in which the queries go in the input.

    Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams, or the %I64d specifier.

    Sample test(s)
    Input
    5 
    4 10 3 13 7
    8
    1 2 4
    2 1 3 3
    1 2 4
    1 3 3
    2 2 5 5
    1 1 5
    2 1 2 10
    1 2 3
    Output
    26 
    22
    0
    34
    11
    Input
    6 
    4 7 4 0 7 3
    5
    2 2 3 8
    1 1 5
    2 3 5 1
    2 4 5 6
    1 2 3
    Output
    38 
    28

    一个用二维线段树操作的异或题,建立20棵线段数 tree[i]表示这n个数第i为在二进制下的情况 更新的时候将每个数 每个位更新 并用延迟符号 求和的时候按位加起来。

      1 #include <iostream>
      2 #include <stdio.h>
      3 #include <algorithm>
      4 #include <string.h>
      5 #include <math.h>
      6 #include <vector>
      7 #include <stack>
      8 using namespace std;
      9 #define ll long long int
     10 ll a[400000][20];
     11 ll b[400000];
     12 void build(int l,int r,int t)
     13 {
     14     int i;
     15     if(l==r)
     16     {
     17         int x;
     18         scanf("%d",&x);
     19         i=0;
     20         while(x)
     21         {
     22             a[t][i++]=x%2;
     23             x/=2;
     24         }
     25         return ;
     26     }
     27     int m=(l+r)>>1;
     28     build(l,m,t<<1);
     29     build(m+1,r,t<<1|1);
     30     for(i=0; i<20; i++)
     31         a[t][i]=a[t<<1][i]+a[t<<1|1][i];
     32 }
     33 void fun(int l,int r,int t)
     34 {
     35     int z=b[t];
     36     int m=(l+r)>>1;
     37     b[t<<1]^=b[t];
     38     b[t<<1|1]^=b[t];
     39     int i=0;
     40     while(z)
     41     {
     42         if(z%2){
     43             a[t<<1][i]=m-l+1-a[t<<1][i];
     44             a[t<<1|1][i]=r-m-a[t<<1|1][i];
     45         }
     46         i++;
     47         z/=2;
     48     }
     49     b[t]=0;
     50 }
     51 void update(int l,int r,int t,int x,int y,int z)
     52 {
     53     int i;
     54     if(l>=x&&r<=y)
     55     {
     56         i=0;
     57         b[t]^=z;
     58         while(z)
     59         {
     60             if(z%2)
     61                 a[t][i]=r-l+1-a[t][i];
     62             i++;
     63             z/=2;
     64         }
     65         return ;
     66     }
     67     if(b[t])
     68         fun(l,r,t);
     69     int m=(l+r)>>1;
     70     if(x<=m)update(l,m,t<<1,x,y,z);
     71     if(y>m)update(m+1,r,t<<1|1,x,y,z);
     72     for(i=0; i<20; i++)
     73         a[t][i]=a[t<<1][i]+a[t<<1|1][i];
     74 }
     75 ll query(int l,int r,int t,int x,int y)
     76 {
     77     if(l>=x&&r<=y)
     78     {
     79        ll sum=0;
     80        for(int i=0;i<20;i++)
     81        sum+=a[t][i]*(1<<i);
     82        return sum;
     83     }
     84     if(b[t])
     85     fun(l,r,t);
     86     int m=(l+r)>>1;
     87     ll sum=0;
     88     if(x<=m)sum+=query(l,m,t<<1,x,y);
     89     if(y>m)sum+=query(m+1,r,t<<1|1,x,y);
     90     for(int i=0; i<20; i++)
     91         a[t][i]=a[t<<1][i]+a[t<<1|1][i];
     92     return sum;
     93 }
     94 int main()
     95 {
     96     memset(a,0,sizeof(a));
     97     memset(b,0,sizeof(b));
     98     int n;
     99     cin>>n;
    100     build(1,n,1);
    101     int m;
    102     cin>>m;
    103     int i,s,x,y,z;
    104     for(i=0; i<m; i++)
    105     {
    106         scanf("%d",&s);
    107         if(s==1)
    108         {
    109             scanf("%d%d",&x,&y);
    110             cout<<query(1,n,1,x,y)<<endl;
    111         }
    112         else
    113         {
    114            scanf("%d%d%d",&x,&y,&z);
    115             update(1,n,1,x,y,z);
    116         }
    117     }
    118 }
    View Code
  • 相关阅读:
    [原][GIS]ARCGIS投影坐标系转换
    [转][osg]探索未知种族之osg类生物【目录】
    [转][osg]探究osg中的程序设计模式【目录】
    [原][资料整理][osg]osgDB文件读取插件,工作机制,支持格式,自定义插件
    [原][landcover]全球地表植被样例图片
    [转]arcgis for server 10.2 下载及安装
    [原]DOM、DEM、landcover,从tms服务发布格式转arcgis、google服务发布格式
    MySQL 数据库最优化设计原则
    MySQL常用存储引擎及如何选择
    Xtrabackup实现Mysql的InnoDB引擎热备份
  • 原文地址:https://www.cnblogs.com/ERKE/p/3256408.html
Copyright © 2011-2022 走看看