USACO Chapter 1 Section 1.2

Posted 2020-09-11 Dark猫

不知道为什么感觉1.2比1.1反而简单不少

 

 1 /*
 2 ID:xiekeyi1
 3 PROG:milk2
 4 LANG:C++
 5  */
 6 #include<bits/stdc++.h>
 7 using namespace std ; 
 8 const int MAXN = 5010;
 9 struct point  
10 {
11     int begin , end ;
12 } a[MAXN] ;
13 
14 
15 bool cmp( struct point a , struct point b )
16 {
17     if( a.begin < b.begin)
18         return true ;
19     else if( a.begin == b.begin && a.end < b.end )
20         return true ;
21     else
22         return false ; 
23 }
24 int main()
25 {
26     freopen("milk2.in","r",stdin);
27     freopen("milk2.out","w",stdout);
28 
29     int n ;
30     cin >> n ;
31     for( int i = 1 ; i <= n ; i++)
32         cin >> a[i].begin >> a[i].end ;
33     sort(a+1,a+1+n,cmp) ; 
34     int ans1= 0 , ans2 = 0 ;
35     int tem_begin = a[1].begin , tem_end= a[1].end;
36     ans1 = max( tem_end - tem_begin , ans1 );  
37     for( int i = 2 ; i <=n ; i++)
38     {
39         if( a[i].begin <= tem_end)
40         {
41             tem_end = max( tem_end , a[i].end)  ;
42         }
43         else 
44         {
45             ans1 = max(ans1 , tem_end - tem_begin ) ;
46             ans2 =max( ans2 , a[i].begin - tem_end) ;
47 
48             tem_begin = a[i].begin;
49             tem_end = a[i].end ;
50         }
51     }
52 
53     cout << ans1 << ‘ ‘ << ans2 << endl ;
54     return 0 ; 
55 }

Milking Cows

 

 

  1 /*
  2 ID:xiekeyi1
  3 PROG:transform
  4 LANG:C++
  5 */
  6 #include<bits/stdc++.h>
  7 using namespace std ;
  8 //#define DEBUG 
  9 const int maxn = 15 ;
 10 char a[maxn][maxn] , b[maxn][maxn] ,  c[maxn][maxn];
 11 //template<typename T>
 12 //void swap( T &a , T &b )
 13 //{
 14 //    T c ;
 15 //    c = a ;
 16 //    a = b ;
 17 //    b = a  ; 
 18 //    return ; 
 19 //}
 20 
 21 bool judge( char a[maxn][maxn] , char b[maxn][maxn] , int &n )
 22 {
 23     for( int i = 1 ; i <= n ; i++)
 24         for( int j = 1 ; j <= n ; j++)
 25             if( a[i][j] != b[i][j] )
 26                 return false ;
 27     return true ;
 28 }
 29 
 30 
 31 int rolate( int &n )
 32 {
 33     for( int i = 1 ; i <= n ; i++)
 34     {
 35         for( int j = 1 ; j <= n ; j++)
 36         {
 37             c[j][n-i+1] = a[i][j] ;
 38         }
 39     }
 40     if( judge( c , b , n ) ) 
 41         return 1 ;
 42 
 43     for( int i = 1  ; i <= n ; i++)
 44     {
 45         for( int j = 1 ; j <= n ; j++)
 46         {
 47             c[n-i+1][n-j+1] = a[i][j] ;
 48         }
 49     }
 50     if( judge ( c , b , n ) ) 
 51         return 2; 
 52 
 53     for( int i = 1 ; i <= n ; i++)
 54     {
 55         for( int j = 1 ; j <= n ; j++)
 56         {
 57             c[n-j+1][i] = a[i][j] ;
 58         }
 59     }
 60 
 61     if( judge ( c , b , n ) )
 62         return 3 ; 
 63     for( int i = 1 ; i <= n ; i++)
 64         for( int j = 1 ; j <= n ; j++)
 65             c[i][j] = a[i][j] ;
 66     for( int i = 1 ; i <= n ; i++)
 67     {
 68         for( int j = 1 ; j <= n / 2 ; j++)
 69         {
 70             swap( c[i][j] , c[i][n-j+1]) ;
 71         }
 72     }
 73     if( judge( c , b , n ) ) 
 74         return 4 ;
 75 
 76     //    for( int i = 1 ; i  <= n ; i++)
 77     //        for( int j = 1 ;  j <=n ; j++)
 78     //            c[i][j] = a[i][j] ;
 79 #ifdef DEBUG
 80     for( int i = 1 ; i <= n ; i++)
 81     {
 82         for( int j = 1 ; j <= n ; j++)
 83             cout << c[i][j];
 84         cout << endl ;
 85     }
 86     cout << endl << endl ; 
 87 #endif 
 88     char d[maxn][maxn];
 89     for( int i = 1 ; i <= n ; i++)
 90         for( int j = 1 ; j <= n ; j++)
 91             d[j][n-i+1] = c[i][j] ;
 92 #ifdef DEBUG
 93     for( int i = 1 ; i <= n ; i++)
 94     {
 95         for( int j = 1 ; j <= n ; j++)
 96             cout << d[i][j] ;
 97         cout << endl ;
 98     }
 99     cout << endl << endl ;
100 #endif 
101     if( judge( d , b , n ) )  
102         return 5 ;
103     for( int i = 1 ; i <= n ; i++)
104         for( int j = 1 ; j <= n ; j++)
105             d[n-i+1][n-j+1] = c[i][j] ;
106 #ifdef DEBUG
107     for( int i = 1 ; i <= n ; i++)
108     {
109         for( int j = 1 ; j <= n ; j++)
110             cout << d[i][j] ;
111         cout << endl ;
112     }
113     cout << endl << endl ;
114 #endif
115     if( judge( d , b , n ) ) 
116         return 5 ;
117     for( int i = 1 ; i <= n ; i++)
118         for( int j = 1 ; j <= n ; j++)
119             d[n-j+1][i] = a[i][j] ;
120 #ifdef DEBUG
121     for( int i = 1 ; i <= n ; i++)
122     {
123         for( int j = 1 ; j <= n ; j++)
124             cout << d[i][j] ;
125         cout << endl ;
126     }
127     cout << endl << endl ;
128 #endif
129     if( judge( d , b , n ) ) 
130         return 5 ; 
131 
132     if( judge( a , b , n ) ) 
133         return 6 ;
134 
135     return 7 ; 
136 
137 }
138 
139 
140 
141 int main()
142 {
143     freopen("transform.in","r",stdin);
144     freopen("transform.out","w",stdout) ; 
145     int n ;
146     cin >> n ;
147     char ch ;
148     for( int i = 1 ; i <= n ; i++)
149         for( int j = 1 ; j <=n ; j++)
150         {
151             cin >> ch ;
152             a[i][j] = ch ;
153         }
154 
155     for( int i = 1 ; i <= n ; i++ )
156         for( int j = 1 ; j <= n ; j++)
157         {
158             cin >> ch ;
159             b[i][j] = ch ; 
160         }
161 
162     cout << rolate(n) << endl ;
163     return 0 ; 
164 }

Transformations

 

 

 1 /*
 2 ID:xiekeyi
 3 PROG:namenum
 4 LANG:C++
 5  */
 6 #include<bits/stdc++.h>
 7 #include<iostream>
 8 using namespace std ;
 9 
10 int f( char ch )
11 {
12     if( ch < ‘Q‘ ) 
13         return ( ch - ‘A‘  + 1 + 2 ) / 3 + 1 ;
14     else
15         return ( ch - ‘A‘ + 1 + 1 ) / 3 + 1 ;
16 
17 }
18 
19 long long func( string &s) 
20 {
21     long long ans = 0 ;
22     for( int i = 0 ; i < s.size() ; i++)
23         ans = ans*10 + f(s[i]) ;
24     return ans ;
25 }
26 
27 int main()
28 {
29     ifstream fin ; 
30     freopen("namenum.out","w",stdout); 
31     fin.open("namenum.in",fstream::in);
32     long long n ;
33     fin >> n ;
34     string s ;
35     fin.close() ; 
36     fin.open("dict.txt",fstream::in);
37     bool flag = false ; 
38     while( fin >> s )
39         if( func(s) == n )
40         {
41             cout << s << endl ;
42             flag = true ; 
43         }
44     if( !flag )
45         cout << "NONE" << endl ; 
46     fin.close() ; 
47     return 0 ; 
48 }

Name That Number

 

 

 

 1 /*
 2 ID:xiekeyi1
 3 PROG:palsquare
 4 LANG:C++
 5  */
 6 #include<bits/stdc++.h>
 7 using namespace std ;
 8 const int maxn = 1000;
 9 int a[maxn] , b[maxn] ;
10 
11 
12 
13 void translate( int n ,  int B , int flag , int &d )
14 {
15     int digit = 0 ;
16     while( n != 0 )
17     {
18         if( flag == 1 )
19             a[digit++] = n % B ;
20         else 
21             b[digit++] = n % B ; 
22         n /= B ;
23     }
24     d = digit ; 
25     return ; 
26 
27 }
28 
29 bool judge( int b[] , int d )
30 {
31     for( int i = 0 ; i <= d / 2 ; i++)
32         if( b[i] != b[d-i-1] )
33             return false ;
34     return true ;
35 }
36 
37 
38 ostream& p( int b[] ,  int d ) 
39 {
40     for( int i = d - 1 ; i >= 0 ; i--)
41     {
42         if( b[i] < 10 )
43             cout << b[i] ;
44         else 
45             cout << static_cast<char> (  b[i] - 10 + ‘A‘ ) ;
46     }
47     return cout ; 
48 }
49 
50 int main()
51 {
52     freopen("palsquare.in","r",stdin);
53     freopen("palsquare.out","w",stdout);
54     int B ;
55     cin >> B ;
56     for( int i = 1 ; i <= 300 ; i++)
57     {
58         int d1 = 0 , d2 = 0 ;
59         int t = i * i ; 
60         translate( t , B , 2 , d2 ) ;
61         if( judge( b , d2 ) )
62         {
63             translate( i , B , 1 , d1 ) ; 
64             p(a,d1) << ‘ ‘ ; 
65             p(b,d2) << endl ; 
66 
67         }
68     }
69 
70     return 0 ; 
71 }

Palindromic Squares

 

 1 /*
 2 ID:xiekeyi1
 3 PROG:dualpal
 4 LANG:C++
 5  */
 6 #include<bits/stdc++.h>
 7 using namespace std ;
 8 const int maxn = 100 ;
 9 int a[maxn] ;
10 
11 
12 void f( int a[] , int n ,  int b , int &d )
13 {
14     int digit = 0 ;
15     while( n != 0 )
16     {
17         a[digit++] = n % b ;
18         n /= b ;
19     }
20     d = digit ;
21     return ; 
22 }
23 
24 bool  judge( int a[] , int d )
25 {
26     for( int i = 0 ; i <= d/2 ; i++)
27         if( a[i] != a[ d-i-1] )
28             return false ;
29     return true ;
30 }
31 int main()
32 {
33 
34     freopen("dualpal.in","r",stdin);
35     freopen("dualpal.out","w",stdout);
36     int n , s ;
37     cin >> n >> s ;
38     int i = 0 ;
39     int temp = s+1 ;
40     for( i = 0 ; i < n ;  ) 
41     {
42         int flag = 0 ; 
43         for( int j = 2 ; j <= 10 ; j++)
44         {
45             int d = 0 ;
46             f(a,temp,j,d);
47             if( judge( a , d ) ) 
48                 flag++;
49             if( flag >= 2 )
50             {
51                 cout << temp << endl ;
52                 i++;
53                 break ; 
54             }
55         }
56         temp++;
57     }
58     return 0 ; 
59 }
60  
61         

Dual Palindromes