###250pt EllysPairs

排序一下求最大最小即可。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std ;
#define For(i , n) for(int i = 0 ; i < (n) ; ++i)
#define SZ(x)  (int)((x).size())
typedef long long lint ;
const int maxint = -1u>>2 ;
const double eps = 1e-6 ; 
class EllysPairs
{
public:
int getDifference(vector <int> knowledge)
{
    int tmp;
    vector <int> g;
    g.clear();
    sort(knowledge.begin(), knowledge.end());
    for(int i = 0; i < SZ(knowledge) / 2; i ++) {
        tmp = knowledge[i] + knowledge[SZ(knowledge) - i - 1];
        g.push_back(tmp);
    }
    sort(g.begin(), g.end());
    return int(g[SZ(g) - 1] - g[0]);
}

};

###500pt EllysFigurines

枚举行或者列的所有操作，然后再去判断列或者行的操作。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
using namespace std ;
#define For(i , n) for(int i = 0 ; i < (n) ; ++i)
#define SZ(x)  (int)((x).size())
typedef long long lint ;
const int maxint = -1u>>2 ;
const double eps = 1e-6 ;

class EllysFigurines
{
public:
 
char maze[40][40];

void init(vector <string> board) {
    for(int i = 0; i < SZ(board); i ++) {
        for(int j = 0; j < SZ(board[0]); j ++) {
            maze[i][j] = board[i][j];
        }
    }
}

int getMoves(vector <string> board, int R, int C) {
    bool flag;
    int m[40]; 
    int ans = 40000, tans, t;
    for(int i = 0; i < 1 << (SZ(board) + 1); i ++) {
        tans = 0;
        init(board);
        t = 0;
        while(t < SZ(board)) {
            int tmp = 1 << t;
            if((i & tmp) > 0) {
                for(int j = t; j < t + R; j ++) {
                    for(int k = 0; k < SZ(board[0]); k ++) {
                        maze[j][k] = '.';
                    }
                }
                t += R;
                tans ++;
            } 
            else {
                t ++;
            }
        }
        memset(m, 0, sizeof(m));
        for(int j = 0; j < SZ(board[0]); j ++) {
            flag = true;
            for(int k = 0; k < SZ(board); k ++) {
                if(maze[k][j] == 'X') {
                    flag = false;
                    break;
                }
            }
            if(!flag) {
                m[j] = 1;
            }
        }
        t = 0;
        while(t < SZ(board[0])) {
            if(m[t]) {
                t += C;
                tans ++;
            }
            else {
                t ++;
            }
        }
        ans = min(tans, ans);
    }
    return (ans);
}

};

###1000pt EllysReversals

每次只能逆转前偶数个元素，就说明每两个字母可以列一组，组内可以随意交换，组间可以随意交换，将每个字符串用字典序排序即可。

注意如果是字母总和如果是奇数，最后一个字母不能改变位置。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std ;
#define For(i , n) for(int i = 0 ; i < (n) ; ++i)
#define SZ(x)  (int)((x).size())
typedef long long lint ;
const int maxint = -1u>>2 ;
const double eps = 1e-6 ; 

class EllysReversals
{
public:
int getMin(vector <string> words)
{
    int ans = 0;
    bool flag[60];
    string tmp;
    vector <string> rec[60];
    memset(flag, true, sizeof(flag));
    for(int i = 0; i < SZ(words); i ++) {
        rec[i].clear();
        for(int j = 0; j < SZ(words[i]); j += 2) { 
            tmp = "";
            tmp += words[i][j];
            if((j + 1) < SZ(words[i]))
                tmp += words[i][j + 1];
            sort(tmp.begin(), tmp.end());
            if(SZ(tmp) == 2)
                rec[i].push_back(tmp);
        }
        sort(rec[i].begin(), rec[i].end());
        if(SZ(words[i]) % 2 != 0) {
            rec[i].push_back(tmp);
        }
        words[i] = "";
        for(int k = 0; k < SZ(rec[i]); k ++) {
            words[i] += rec[i][k];
        }
    }
    for(int i = 0; i < SZ(words); i ++) {
        for(int j = i + 1; j < SZ(words); j ++) {
            if(words[i] == words[j] && flag[i] && flag[j]) {
                flag[i] = flag[j] = false;
                ans ++;
            }
        }
    }
    ans = SZ(words) - ans * 2;
    return int(ans);
}

};