Codeforces Round #234 (Div. 2) 解题报告

题意：水题

代码如下：

 1 /**************************************************
 2  * Author     : xiaohao Z
 3  * Blog     : http://www.cnblogs.com/shu-xiaohao/
 4  * Last modified : 2014-03-05 23:13
 5  * Filename     : Codeforce_234_2_A.cpp
 6  * Description     : 
 7  * ************************************************/
 8 
 9 #include <iostream>
10 #include <cstdio>
11 #include <cstring>
12 #include <cstdlib>
13 #include <cmath>
14 #include <algorithm>
15 #include <queue>
16 #include <stack>
17 #include <vector>
18 #include <set>
19 #include <map>
20 #define MP(a, b) make_pair(a, b)
21 #define PB(a) push_back(a)
22 
23 using namespace std;
24 typedef long long ll;
25 typedef pair<int, int> pii;
26 typedef pair<unsigned int,unsigned int> puu;
27 typedef pair<int, double> pid;
28 typedef pair<ll, int> pli;
29 typedef pair<int, ll> pil;
30 
31 const int INF = 0x3f3f3f3f;
32 const double eps = 1E-6;
33 int mp[20][20];
34 vector<pii> p;
35 string str;
36 
37 void debug(int a, int b){
38     for(int i=0; i<a; i++){
39         for(int j=0; j<b; j++){
40             cout << mp[i][j] << ' ';
41         }cout << endl;
42     }cout << endl;
43 }
44 
45 bool calc(int a, int b){
46     for(int i=0; i<a; i++){
47         for(int j=0; j<b; j++){
48             if(str[i*b+j] == 'X') mp[i][j] = 1;
49             else mp[i][j] = 0;
50         }
51     }
52     for(int i=0; i<b; i++){
53         int f = 1;
54         for(int j=0; j<a; j++){
55             if(mp[j][i] == 0){
56                 f = 0;
57                break;    
58             }    
59         }
60         if(f) return true;
61     }
62     return false;
63 }
64 
65 int main()
66 {
67 //    freopen("in.txt", "r", stdin);
68 
69     int t;
70     cin >> t;
71     while(t--){
72         cin >> str;
73         int ans = 0;
74         p.clear();
75         for(int i=1; i<=12; i++){
76             for(int j=1; j<=12; j++){
77                 if(i*j==12){
78                     if(calc(i, j)) {
79                         p.PB(MP(i, j));
80                         ans++;     
81                     }
82                 }
83             }
84         }
85         cout << ans;
86         for(int i=0; i<ans; i++){
87             cout << ' ' << p[i].first << 'x' << p[i].second;
88         }
89         cout << endl;
90     }
91     return 0;
92 }

View Code

Problem B Inna and New Matrix of Candies
思路：每次找一个最小的然后全部向右移动，计算次数。

代码如下：

 1 /**************************************************
 2  * Author     : xiaohao Z
 3  * Blog     : http://www.cnblogs.com/shu-xiaohao/
 4  * Last modified : 2014-03-05 23:15
 5  * Filename     : Codeforce_234_2_B.cpp
 6  * Description     : 
 7  * ************************************************/
 8 
 9 #include <iostream>
10 #include <cstdio>
11 #include <cstring>
12 #include <cstdlib>
13 #include <cmath>
14 #include <algorithm>
15 #include <queue>
16 #include <stack>
17 #include <vector>
18 #include <set>
19 #include <map>
20 #define MP(a, b) make_pair(a, b)
21 #define PB(a) push_back(a)
22 
23 using namespace std;
24 typedef long long ll;
25 typedef pair<int, int> pii;
26 typedef pair<unsigned int,unsigned int> puu;
27 typedef pair<int, double> pid;
28 typedef pair<ll, int> pli;
29 typedef pair<int, ll> pil;
30 
31 const int INF = 0x3f3f3f3f;
32 const double eps = 1E-6;
33 const int LEN = 1010;
34 int Map[LEN][LEN], n, m;
35 vector<int> p;
36 
37 int main()
38 {
39 //    freopen("in.txt", "r", stdin);
40 
41     string str;
42     int ans = 1, pos;
43     while(cin >> n >> m){
44         ans = 1;
45         p.clear();
46         for(int i=0; i<n; i++){
47             cin >> str;
48             pos = -1;
49             for(int j=0; j<m; j++){
50                 if(str[j] == 'G') pos = j;
51                 if(pos == -1 && str[j] == 'S') ans = 0;
52                 if(str[j] == 'S') p.PB(j-pos);
53             }
54         }
55         sort(p.begin(), p.end());
56     //    for(int i=0; i<p.size(); i++)cout << p[i] << ' ';cout << endl;
57         int cnt = 0;
58         for(int i=0; i<p.size(); i++){
59             if(p[i] == 0) continue;
60             cnt ++;
61             for(int j=p.size()-1; j>i; j--){
62                 p[j] -= p[i];
63             }
64         }
65         if(ans) cout << cnt;
66         else cout << "-1";
67         cout << endl;
68     }
69     return 0;
70 }

View Code

Problem C Inna and Huge Candy Matrix

题意：一个矩阵中有糖果，告诉你糖果的位置，然后先顺时针转90度x次，镜像y次，逆时针90度z次问你糖果的位置。

思路：直接模拟，注意每次旋转交换n和m

代码如下：

 1 /**************************************************
 2  * Author     : xiaohao Z
 3  * Blog     : http://www.cnblogs.com/shu-xiaohao/
 4  * Last modified : 2014-03-05 23:14
 5  * Filename     : Codeforce_234_2_C.cpp
 6  * Description     : 
 7  * ************************************************/
 8 
 9 #include <iostream>
10 #include <cstdio>
11 #include <cstring>
12 #include <cstdlib>
13 #include <cmath>
14 #include <algorithm>
15 #include <queue>
16 #include <stack>
17 #include <vector>
18 #include <set>
19 #include <map>
20 #define MP(a, b) make_pair(a, b)
21 #define PB(a) push_back(a)
22 
23 using namespace std;
24 typedef long long ll;
25 typedef pair<int, int> pii;
26 typedef pair<unsigned int,unsigned int> puu;
27 typedef pair<int, double> pid;
28 typedef pair<ll, int> pli;
29 typedef pair<int, ll> pil;
30 
31 const int INF = 0x3f3f3f3f;
32 const double eps = 1E-6;
33 int n, m, x, y, z, p, a, b;
34 
35 pii ca(pii t, int x){
36     pii ret;
37     while(x--){
38         ret = t;
39         ret.first = t.second;
40         ret.second = n - t.first + 1;
41         t = ret;
42         swap(n, m);
43     }
44     return t;
45 }
46 
47 pii cb(pii t, int x){
48     pii ret = t;
49     while(x--){
50         ret = t;
51         ret.second = m - t.second + 1;
52     }
53     return ret;
54 }
55 
56 pii cc(pii t, int x){
57     pii ret;
58     while(x--){
59         ret = t;
60         ret.second = t.first;
61         ret.first = m - t.second + 1;
62         t = ret;
63         swap(n, m);
64     }
65     return t;
66 }
67 
68 int main()
69 {
70 //    freopen("in.txt", "r", stdin);
71 
72     pii sw;
73     int tn, tm;
74     while(cin >> tn >> tm >> x >> y >> z >> p){
75         for(int i=0; i<p; i++){
76             cin >> a >> b;
77             n = tn, m = tm;
78             sw = MP(a, b);
79             sw = ca(sw, x%4);
80             sw = cb(sw, y%2);
81             sw = cc(sw, z%4);
82             cout << sw.first << ' ' << sw.second << endl;
83         }
84     }
85     return 0;
86 }

View Code

奔跑吧！少年！趁着你还年轻