C - 2D Plane 2N Points

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Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (ai,bi), and the coordinates of the i-th blue point are (ci,di).

A red point and a blue point can form a friendly pair when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point.

At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.

Constraints

• All input values are integers.
• 1≤N≤100
• 0≤ai,bi,ci,di<2N
• a1,a2,…,aN,c1,c2,…,cN are all different.
• b1,b2,…,bN,d1,d2,…,dN are all different.

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Input

Input is given from Standard Input in the following format:

Na1 b1a2 b2:aN bNc1 d1c2 d2:cN dN

Output

Print the maximum number of friendly pairs.

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Sample Input 1

Copy

32 03 11 34 20 45 5

Sample Output 1

Copy

2

For example, you can pair (2,0) and (4,2), then (3,1) and (5,5).

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Sample Input 2

Copy

30 01 15 22 33 44 5

Sample Output 2

Copy

2

For example, you can pair (0,0) and (2,3), then (1,1) and (3,4).

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Sample Input 3

Copy

22 23 30 01 1

Sample Output 3

Copy

0

It is possible that no pair can be formed.

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Sample Input 4

Copy

50 07 32 24 81 68 56 95 49 13 7

Sample Output 4

Copy

5

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Sample Input 5

Copy

50 01 15 56 67 72 23 34 48 89 9

Sample Output 5

Copy

4

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这个题目的描述很简单，就是有A，B两个集合，你要找到两个集合最多匹配的个数，匹配的是B中的元素横纵坐标都是比A的大，这里采用二分，查询另一个分组是不是存在这样的元素，每次都选择与其最相近的元素

（相当于排序之后不够考虑一个坐标轴）

#include
     
      using namespace std;#define Fi first#define Se secondpair
      
       
         ,int>p[1005];int n;int main(){    cin>>n;    for(int i=0; i
        
         >p[i].Fi.Fi>>p[i].Fi.Se,p[i].Se=i>=n;    sort(p,p+n+n);    set
         
           s;    int ans=0;    for(int i=0; i
          
           p[i].Fi.Se))s.erase(--s.lower_bound(p[i].Fi.Se)),ans++; else if(!p[i].Se)s.insert(p[i].Fi.Se); cout<
          
         
        
       
      
     

D - Two Sequences

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Time limit : 3sec / Memory limit : 256MB

Score : 500 points

Problem Statement

You are given two integer sequences, each of length N: a1,…,aN and b1,…,bN.

There are N2 ways to choose two integers i and j such that 1≤i,j≤N. For each of these N2 pairs, we will compute ai+bj and write it on a sheet of paper. That is, we will write N2 integers in total.

Compute the XOR of these N2 integers.

 

Definition of XOR

 

Constraints

• All input values are integers.
• 1≤N≤200,000
• 0≤ai,bi<228

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Input

Input is given from Standard Input in the following format:

Na1 a2 … aNb1 b2 … bN

Output

Print the result of the computation.

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Sample Input 1

Copy

21 23 4

Sample Output 1

Copy

2

On the sheet, the following four integers will be written: 4(1+3),5(1+4),5(2+3) and 6(2+4).

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Sample Input 2

Copy

64 6 0 0 3 30 5 6 5 0 3

Sample Output 2

Copy

8

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Sample Input 3

Copy

51 2 3 4 51 2 3 4 5

Sample Output 3

Copy

2

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Sample Input 4

Copy

100

Sample Output 4

Copy

0

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这个题目也是有点意思，题目还是很好懂的

 

就是给你A数组和B组数，A和B进行相加有n*n个结果，求这n*n结果的异或和

首先你要想n*n的运算肯定是爆炸的

a+b = a^b + (a&b)<<1

所以n为a次表示要进行n次异或，异或两次的值为0，所以n为奇数答案肯定有两个数组的异或和

然后将其划分为子问题，选取两个数的和最大值2^29

用位运算实现两数相加

int Add(int a,int b){    return b?Add(a^b,(a&b)<<1):a;}

 位运算实现两数相减

int MINUS(int a,int b){    return b?MINUS(a^b,((a^b)&b)<<1):a;}

 

#include
       
        using namespace std;const int N=2e5+5;int a[N],b[N];int main(){    int n,r=0;    cin>>n;    for(int i=0; i
        
         >a[i];    for(int i=0; i
         
          >b[i];    if(n&1)    {        for(int i=0; i
          
           >=1)    {        for(int j=0; j