sort題目

uva,146

全排列問題：permutation 

具體詳解：參考Devymex

UVa Problem 146 - ID Codes

 

Problem:

Please find the problem  here.

Solution:

This is simply the next permutation problem. Thanks to the hint on Competitive Programming, there is a simple call in C++ library named next_permutation that can solve the problem, and therefore I used it.

Suppose we are not given the next_permutation code, how would I solve it? First, note that in order to find the least significant digit to increase, but you can't just increase a digit because this is a permutation, one also need to change another digit to change as well. Surely we don't want to alter the more significant digits, so we can only pick the less significant ones.

Formally, one scan from right to left, trying to find a digit such that there is a larger digit to its right. If such a digit does not exist, we can conclude there is no successor. Otherwise, change that digit to the least larger one found on the right, and sort the rest of the digits, that gives the answer.

For example, the successor of 1, 2, 3, 3 is 1, 3, 2, 3, this is done by noting 2 < 3, so we switch 2 to 3, and just sort [2, 3], which give [2, 3]

As a more exotic scenario, let's look at 1, 4, 3, 2, its successor is 2, 1, 3, 4, this is done by noting 1 < 2, so we switch 1 to 2, and just sort [1, 4, 3], which gives [1, 3, 4].

 

貼一下本身寫的代碼：

老是wrong answer不知道怎麼的：到時候再來填坑。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <vector>
 5 
 6 using namespace std;
 7 
 8 int main()
 9 {
10     vector<char> s;
11     while(!cin.eof())
12     {
13         char c=cin.get();
14         if(c!=-1)
15         {
16             if(c!='\r' && c!='\n')
17             {
18                 if(c=='#')
19                     break;
20                 s.push_back(c);
21             }
22             else
23             {
24                 vector<char>::iterator it=s.end(),at;
25                 for(--it;it!=s.begin();--it)
26                 {
27                     if(*(it-1)<*it)
28                         break;
29                 }
30                 if(it==s.begin())
31                     printf("No Successor\n");
32                 else
33                 {
34                     it--;
35                     for(at=it+1;at!=s.end();at++)
36                         if(*at<*it)
37                             break;
38                     iter_swap(it,--at);
39                     reverse(it+1,s.end());
40                     for(vector<char>::iterator i=s.begin();i!=s.end();i++)
41                         cout<<*i;
42                     cout<<endl;
43                 }
44                 s.clear();
45                 
46             }
47         }
48     }
49     return 0;
50 }

View Code

accept代碼：Andrew

 1 #include <iostream>
 2 #include <vector>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 
 7 int main()
 8 {
 9     vector<char> line;
10     while (!cin.eof())
11     {   
12         char c = cin.get();
13         if (c != -1)
14         {
15             if (c != '\r' && c != '\n')
16             {
17                 if (c == '#')
18                 {
19                     break;
20                 }
21                 line.push_back(c);
22             }
23             else
24             {
25                 if (next_permutation(line.begin(), line.end()))
26                 {
27                     for (vector<char>::iterator i = line.begin(); i != line.end(); i++)
28                     {
29                         cout << *i;
30                     }
31                     cout << endl;
32                 }
33                 else
34                 {
35                     cout << "No Successor" << endl;
36                 }
37                 line.clear();
38             }
39         }
40     }
41 
42     return 0;
43 }

View Code

uva,299

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 const int maxn=55;
 7 
 8 int main()
 9 {
10     int N;
11     int L;
12     int a[maxn];
13     scanf("%d",&N);
14     while(N--)
15     {
16         scanf("%d",&L);
17         for(int i=0;i<L;i++)
18             scanf("%d",&a[i]);
19         int count=0;
20         for(int i=0;i<L;i++)
21             for(int j=1;j<L-i;j++)
22             {
23                 if(a[j]<a[j-1])
24                 {
25                     swap(a[j],a[j-1]);
26                     count++;
27                 }
28             }
29         printf("Optimal train swapping takes %d swaps.\n",count);
30     }
31     return 0;
32 }

View Code

======

用結構體來模擬map很舒服。

uva,10008

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdio>
 4 #include <cstring>
 5 
 6 using namespace std;
 7 
 8 struct P
 9 {
10     int times;
11     char ch;
12 };
13 
14 int comp(const P x,const P y)
15 {
16     if(x.times==y.times)
17         return x.ch<y.ch;
18     return x.times>y.times;
19 }
20 int main()
21 {
22     int n;
23     char ch,s[10000];
24     P a[27];
25     for(int i=0;i<26;i++)
26     {
27         a[i].times=0;
28         a[i].ch=i+65;
29     }
30     scanf("%d",&n);
31     while(n--)
32     {
33         cin.getline(s,10000);
34         for(int i=0;i<strlen(s);i++)
35         {
36             if(isalpha(s[i]))
37             {
38                 if(islower(s[i])) s[i]-=32;
39                 a[(int)s[i]-65].times++;
40             }
41         }
42     }
43     sort(a,a+26,comp);
44     for(int i=0;i<26,a[i].times>0;i++)
45         cout<<a[i].ch<<" "<<a[i].times<<endl;
46     
47 }

View Code

uva,10041

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <math.h>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 int main()
 8 {
 9     int t,r;
10     int s[500];
11     scanf("%d",&t);
12     while(t--)
13     {
14         scanf("%d",&r);
15         for(int i=0;i<r;i++)
16             scanf("%d",&s[i]);
17         sort(s,s+r);
18         int count=s[r/2];
19         int total=0;
20         int t=r/2;
21         for(int i=0;i<r;i++)
22         {
23             if(i==t)
24                 continue;
25             total+=abs(count-s[i]);
26         }
27         printf("%d\n",total);
28     }
29     return 0;
30 }

View Code

uva,10050

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 const int maxn=3655;
 6 
 7 int main()
 8 {
 9     int T,P,N;
10     int day[maxn];
11     int h[105];
12     int k;
13     scanf("%d",&T);
14     while(T--)
15     {
16         scanf("%d",&N);
17         scanf("%d",&P);
18         for(int i=0;i<P;i++)
19         {
20             scanf("%d",&h[i]);
21         }
22         memset(day,0,sizeof(day));
23         for(int i=0;i<P;i++)
24         {
25             k=1;
26             while(k*h[i]<=N)
27             {
28                 day[k*h[i]]=1;
29                 k++;
30             }
31         }
32         int count=0;
33             for(int i=1;i<=N;i++)
34             {
35                 if(i%7==6 || i%7==0)
36                     continue;
37                 if(day[i]==1)
38                     count++;
39             }
40         printf("%d\n",count);
41     }
42     return 0;
43 }

View Code

uva,10107

 1 #include <iostream>
 2 #include <algorithm>
 3 using namespace std;
 4 
 5 int main()
 6 {
 7     long long a[10005],l,X;
 8     int i=0;
 9     while(cin>>X)
10     {
11         a[i]=X;
12         sort(a,a+i+1);
13         if(i%2==0)
14             cout<<a[i/2]<<endl;
15         else
16         {
17             l=a[(i-1)/2]+a[(i+1)/2];
18             cout<<l/2<<endl;
19         }
20         i++;
21     }
22     return 0;
23 }

View Code

uva,10327

 1 #include <iostream>
 2 #include <cstdio>
 3 
 4 using namespace std;
 5 const int maxn=1005;
 6 
 7 void flip(int *x,int *y)
 8 {
 9     int tmp;
10     tmp=*x;
11     *x=*y;
12     *y=tmp;
13 }
14 
15 int main()
16 {
17     int N;
18     int a[maxn];
19     while((scanf("%d",&N))!=EOF)
20     {
21         for(int i=0;i<N;i++)
22             scanf("%d",&a[i]);
23         int count=0;
24         for(int i=0;i<N;i++)
25         {
26             for(int j=1;j<N-i;j++)
27             {
28                 if(a[j]<a[j-1])
29                 {
30                     flip(&a[j],&a[j-1]);
31                     count++;
32                 }
33             }
34         }
35         printf("Minimum exchange operations : %d\n",count);
36     }    
37     return 0;
38 }

View Code