Contest - 2014 SWJTU ACM 手速测试赛（2014.10.31）

题目列表：

2146 Problem A	【手速】阔绰的Dim
2147 Problem B	【手速】颓废的Dim
2148 Problem C	【手速】我的滑板鞋
2149 Problem D	【手速】潦倒的Dim
2150 Problem E	【手速】被NTR的Dim

2146 Problem A：

简单的最长回文串统计算法，这里没有过高要求，n^2算法可以AC。其中包括dp动规以及中心法（以上两种都是O(n^2)算法，可以参考白书）。推广，可以尝试扩展KMP(O(nlogn))或者Manacher算法（O(n)）。可以查阅相关资料自行选择学习。这里给出中心法。

 1 /****************************************/
 2 /*****            Desgard_Duan        *****/
 3 /****************************************/
 4 //#pragma comment(linker, "/STACK:102400000,102400000")
 5 #define _CRT_SECURE_NO_WARNINGS
 6 #include <iostream>
 7 #include <cstdio>
 8 #include <cstdlib>
 9 #include <cstring>
10 #include <string>
11 #include <algorithm>
12 #include <stack>
13 #include <map>
14 #include <queue>
15 #include <vector>
16 #include <set>
17 #include <functional>
18 #include <cmath>
19 #include <numeric>
20 
21 using namespace std;
22 
23 char s[500]; int len;
24 
25 inline void get_val(int &a) {
26     int value = 0, s = 1;
27     char c;
28     while ((c = getchar()) == ' ' || c == '\n');
29     if (c == '-') s = -s; else value = c - 48;
30     while ((c = getchar()) >= '0' && c <= '9')
31         value = value * 10 + c - 48;
32     a = s * value;
33 }
34 
35 int tofind(int h, int t) {
36     int re = t - h + 1;
37     while (h < t) {
38         if (s[h++] != s[t--]) return 0;
39     }
40     return re;
41 }
42 
43 int main() {
44     //freopen("huiwen.in", "r", stdin);
45     //freopen("huiwen.out","w",stdout);
46     int T;
47     cin >> T;
48     while (T--) {
49         scanf("%s", s);
50         len = strlen(s);
51         int ans = 1;
52         for (int i = 0; i < len - 1; i++)
53             for (int j = i + 1; j < len; j++) {
54                 ans = max(ans, tofind(i, j));
55             }
56         cout << ans << endl;
57     }
58     return 0;
59 }

2147 Problem B:

一道字符串水题，只要按位置遍历一遍即可。C语言基础题。

 1 /****************************************/
 2 /*****            Desgard_Duan        *****/
 3 /****************************************/
 4 //#pragma comment(linker, "/STACK:102400000,102400000")
 5 #define _CRT_SECURE_NO_WARNINGS
 6 #include <iostream>
 7 #include <cstdio>
 8 #include <cstdlib>
 9 #include <cstring>
10 #include <string>
11 #include <algorithm>
12 #include <stack>
13 #include <map>
14 #include <queue>
15 #include <vector>
16 #include <set>
17 #include <functional>
18 #include <cmath>
19 #include <numeric>
20 
21 using namespace std;
22 
23 inline void get_val(int &a) {
24     int value = 0, s = 1;
25     char c;
26     while ((c = getchar()) == ' ' || c == '\n');
27     if (c == '-') s = -s;
28     else value = c - 48;
29     while ((c = getchar()) >= '0' && c <= '9')
30         value = value * 10 + c - 48;
31     a = s * value;
32 }
33 
34 string str1, str2;
35 int main () {
36     int T;
37     //cin >> T;
38     while (cin >> str1 >> str2) {
39 
40         int ans = 0;
41         for (int i = 0 ; i < str1.size(); ++ i) {
42             if (str1[i] == str2[i]) {
43                 ans ++;
44             }
45         }
46         cout << ans << endl;
47     }
48     return 0;
49 }

2148 Problem C:

一道贪心的白书例题，类型归类为查找不相交区间的最大个数。具体思路：对于相交的任意两个区间分为两种情况（图A、图B）。若出现情况A，直接将大区间删除即可。若出现情况B，我们先将集合按照x进行升序排列，然后优先选取x最小的情况B中的区间，这样可以得到最佳的方案。

 1 /****************************************/
 2 /*****            Desgard_Duan        *****/
 3 /****************************************/
 4 //#pragma comment(linker, "/STACK:102400000,102400000")
 5 #define _CRT_SECURE_NO_WARNINGS
 6 #include <iostream>
 7 #include <cstdio>
 8 #include <cstdlib>
 9 #include <cstring>
10 #include <string>
11 #include <algorithm>
12 #include <stack>
13 #include <map>
14 #include <queue>
15 #include <vector>
16 #include <set>
17 #include <functional>
18 #include <cmath>
19 #include <numeric>
20 
21 using namespace std;
22 
23 inline void get_val(int &a) {
24     int value = 0, s = 1;
25     char c;
26     while ((c = getchar()) == ' ' || c == '\n');
27     if (c == '-') s = -s;
28     else value = c - 48;
29     while ((c = getchar()) >= '0' && c <= '9')
30         value = value * 10 + c - 48;
31     a = s * value;
32 }
33 
34 vector<pair<int, int> > shoes;
35 bool flag[1005];
36 
37 int main () {
38     int n, x, y;
39     //freopen("out.txt", "w", stdout);
40     while (~scanf ("%d", &n)) {
41         shoes.clear();
42         for (int i = 0; i < n; ++ i) {
43             cin >> x >> y;
44             shoes.push_back (make_pair(x, y));
45         }
46         sort (shoes.begin(), shoes.end());
47 
48         memset (flag, 0, sizeof (flag));
49         for (int i = 0; i < n; ++ i) {
50             if (flag[i]) {
51                 continue;
52             }
53             for (int j = 0; j < n; ++ j) {
54                 if (i == j) continue;
55                 else {
56                     if (shoes[i].first <= shoes[j].first && shoes[i].second >= shoes[j].second) {
57                         flag[i] = 1;
58                     }
59                 }
60             }
61         }
62         int cur = 0, ans = 1;
63         for (; cur < shoes.size() && flag[cur]; cur ++);
64         int last_end = shoes[cur].second, this_begin;
65         for (int i = cur + 1; i < shoes.size(); ++ i) {
66             if (flag[i]) continue;
67             this_begin = shoes[i].first;
68             if (last_end <= this_begin) {
69                 ans ++;
70                 last_end = shoes[i].second;
71             }
72         }
73         cout << ans << endl;
74 
75     }
76     return 0;
77 }

2149 Problem D:

一道大数题目。在n个大数中寻找最小的数。大数推荐使用Java大数类，相对来说代码比较清晰。也可以直接开一个数组进行模拟。

 1 import java.math.*;
 2 import java.io.*;
 3 import java.util.*;
 4 
 5 public class Main {
 6     public static void main(String args[]) {
 7         Scanner in = new Scanner(System.in);
 8         while (in.hasNext()) {
 9             int n = in.nextInt();
10             BigInteger ans = BigInteger.ZERO;
11             for (int i = 0; i < n; ++i) {
12                 BigInteger a = in.nextBigInteger();
13                 if (i == 0) {
14                     ans = a;
15                 } else {
16                     ans = ans.min(a);
17                 }
18             }
19             System.out.println (ans);
20         }
21     }
22 }

2150 Problem E:

一道简单的数学题目。稍微推导一下就会发现这个函数最多只有六项。分别是a, b, b - a, -a, -b, a - b六个数，只要我们去一下重复的数即可。去重方法可以用一个字符串数组来做，这里用了set容器的性质进行了去重操作。

 1 /****************************************/
 2 /*****            Desgard_Duan        *****/
 3 /****************************************/
 4 //#pragma comment(linker, "/STACK:102400000,102400000")
 5 #define _CRT_SECURE_NO_WARNINGS
 6 #include <iostream>
 7 #include <cstdio>
 8 #include <cstdlib>
 9 #include <cstring>
10 #include <string>
11 #include <algorithm>
12 #include <stack>
13 #include <map>
14 #include <queue>
15 #include <vector>
16 #include <set>
17 #include <functional>
18 #include <cmath>
19 #include <numeric>
20 
21 using namespace std;
22 
23 inline void get_val(int &a) {
24     int value = 0, s = 1;
25     char c;
26     while ((c = getchar()) == ' ' || c == '\n');
27     if (c == '-') s = -s;
28     else value = c - 48;
29     while ((c = getchar()) >= '0' && c <= '9')
30         value = value * 10 + c - 48;
31     a = s * value;
32 }
33 
34 
35 int a, b, n;
36 set<int> S;
37 int main () {
38     while (cin >> a >> b >> n) {
39         S.clear();
40         if (n >= 6) {
41             S.insert (a);
42             S.insert (b);
43             S.insert (b - a);
44             S.insert (-a);
45             S.insert (-b);
46             S.insert (a - b);
47             cout << S.size() << endl;
48             continue;
49         } else {
50             int last = a, now = b, t;
51             S.insert (a);
52             S.insert (b);
53             for (int i = 3; i <= n; ++ i) {
54                 S.insert (now - last);
55                 t = now;
56                 now = now - last;
57                 last = t;
58             }
59             cout << S.size() << endl;
60         }
61     }
62     return 0;
63 }