（TOJ1435）{A} + {B}

描述

给你两个集合，要求{A} + {B}. 注:同一个集合中不会有两个相同的元素.

输入

每组输入数据分为三行,第一行有两个数字n,m(0<n,m<=10000),分别表示集合A和集合B的元素个数.后两行分别表示集合A和集合B.每个元素为不超出int范围的整数,每个元素之间有一个空格隔开

输出

针对每组数据输出一行数据,表示合并后的集合,要求从小到大输出,每个元素之间有一个空格隔开

样例输入

1 2

1

2 3

1 2

1

1 2

样例输出

1 2 3

1 2

AC code1：

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<ctype.h>

 4 #include<math.h>

 5 

 6 int a[10001],b[10001];

 7 

 8 int split(int a[],int low,int high)

 9 {

10    int part_element=a[low];

11    for(;;){

12    while(low<high&&part_element<=a[high])

13      high--;

14    if(low>=high) break;

15    a[low++]=a[high];

16  

17    while(low<high&&a[low]<=part_element)

18      low++;

19    if(low>=high) break;

20    a[high--]=a[low];

21    }

22    a[high]=part_element;

23    return high;

24 }

25 void quicksort(int a[],int low,int high)

26 {

27    int middle;

28    if(low>=high)return;

29    middle=split(a,low,high);

30    quicksort(a,low,middle-1);

31    quicksort(a,middle+1,high);

32  }

33 int deal(int c[], int n, int d[], int m)

34 {

35     int L,i,j,flag=0;

36     i=j=0;

37     while(i<n && j<m){

38         if(c[i]<d[j]){

39             if(flag)  printf(" ");

40             printf("%d",c[i]);

41             while(c[i+1]==c[i] && i+1<n) i++;

42             i++;flag=1;

43         }else if(c[i]==d[j]){

44             if(flag) printf(" ");

45             printf("%d",c[i]);

46             while(c[i+1]==c[i] && i+1<n) i++;

47             while(d[j+1]==d[j] && j+1<m) j++;

48             i++; j++; flag=1;

49         }else{

50             if(flag) printf(" ");

51             printf("%d",d[j]);

52             while(d[j+1]==d[j] && j+1<m) j++;

53             j++; flag=1;

54         }

55     }

56     if(i==n){

57         while(j<m)  

58         {

59            printf(" %d",d[j]);

60            while(d[j]==d[j+1]) j++;

61            j++;

62        }

63     }else{

64         while(i<n)

65         {

66           printf(" %d",c[i]);

67             while(c[i]==c[i+1]) i++;

68           i++;

69         }

70      }

71     printf("\n");

72 }

73 

74 void solve()

75 {

76     int i,j,n,m;

77     while(scanf("%d %d",&n,&m)!=EOF){

78     for(i=0; i<n; i++){

79         scanf("%d",&a[i]);

80     }

81     for(i=0; i<m; i++){

82         scanf("%d",&b[i]);

83     }

84     quicksort(a,0,n-1);

85     quicksort(b,0,m-1);

86     deal(a,n,b,m);

87     }

88 }

89 

90 int main()

91 {

92     solve();

93     return 0;

94 }

AC code2：

    

#include<stdio.h>

#include<string.h>

#include<ctype.h>

#include<math.h>



int a[20001];



int split(int a[],int low,int high)

{

   int part_element=a[low];

   for(;;){

   while(low<high&&part_element<=a[high])

     high--;

   if(low>=high) break;

   a[low++]=a[high];

   while(low<high&&a[low]<=part_element)

     low++;

   if(low>=high) break;

   a[high--]=a[low];

   }

   a[high]=part_element;

   return high;

}



void quicksort(int a[],int low,int high)

{

   int middle;

   if(low>=high)return;

   middle=split(a,low,high);

   quicksort(a,low,middle-1);

   quicksort(a,middle+1,high);

}



void deal(int a[], int n)

{

    int i;

    printf("%d",a[0]);

    for(i=1; i<n; i++){

        if(a[i]!=a[i-1])

          printf(" %d",a[i]);

    }

    printf("\n");

} 



void solve()

{

    int i,j,n,m;

    while(scanf("%d %d",&n,&m)!=EOF){

    for(i=0; i<n+m; i++){

        scanf("%d",&a[i]);

    }

    quicksort(a,0,n+m-1);

    deal(a,n+m);

    }

}



int main()

{

    solve();

    return 0;

}

 

C++ code:

 1 #include<iostream>

 2 #include<vector>

 3 #include<iterator>

 4 #include <utility>

 5 #include <algorithm>

 6 using namespace std;

 7 

 8  int main()

 9  {

10      int m,n,i,t;

11      while(cin>>n>>m)

12      {

13         vector<int>v(m+n);

14         vector<int>::iterator it;

15         for(it=v.begin(); it!=v.end(); ++it)

16         {

17             cin>>*it;

18         }

19         sort(v.begin(),v.end(),less<int>());

20         v.erase(unique(v.begin(),v.end()),v.end());

21         for(it=v.begin(); it!=v.end(); ++it)

22         {

23            if(it>v.begin())

24               cout<<" ";

25            cout<<*it;

26         }

27         cout<<endl;

28      }

29     return 0;

30  }