Codeforces Round #490 (Div. 3)

Posted 2021-02-16 acgoto

A、Mishka and Contest

思路：简单贪心。每次删掉首尾不超过k的元素，直到分别第一次遇到超过k的元素就不再继续删除即可。

AC代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <iomanip>
 8 #include <complex>
 9 #include <string>
10 #include <vector>
11 #include <set>
12 #include <map>
13 #include <list>
14 #include <deque>
15 #include <queue>
16 #include <stack>
17 #include <bitset>
18 using namespace std;
19 typedef long long LL;
20 typedef unsigned long long ULL;
21 const int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 上右下左
22 const int mx[8] = {-1, -2, -2, -1, 1, 2, 2, 1}; // 马可走的八个方向
23 const int my[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
24 const double eps = 1e-6;
25 const double PI = acos(-1.0);
26 const int maxn = 105;
27 const int inf = 0x3f3f3f3f;
28 
29 int n, k, ans, a[maxn];
30 
31 int main() {
32     while(cin >> n >> k) {
33         ans = 0;
34         for(int i = 0; i < n; ++i) cin >> a[i];
35         for(int i = 0, j = n - 1; i <= j;) {
36             if(a[i] <= k) ++i, ++ans;
37             else if(a[j] <= k) --j, ++ans;
38             else break;
39         }
40         cout << ans << endl;
41     }
42     return 0;
43 }

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B、Reversing Encryption

思路：暴力。简单按照n的所有公约数从小到大有序反转一下即可。

AC代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <iomanip>
 8 #include <complex>
 9 #include <string>
10 #include <vector>
11 #include <set>
12 #include <map>
13 #include <list>
14 #include <deque>
15 #include <queue>
16 #include <stack>
17 #include <bitset>
18 using namespace std;
19 typedef long long LL;
20 typedef unsigned long long ULL;
21 const int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 上右下左
22 const int mx[8] = {-1, -2, -2, -1, 1, 2, 2, 1}; // 马可走的八个方向
23 const int my[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
24 const double eps = 1e-6;
25 const double PI = acos(-1.0);
26 const int maxn = 1e5+5;
27 const int inf = 0x3f3f3f3f;
28 
29 int n;
30 string str;
31 
32 int main() {
33     while(cin >> n >> str) {
34         for(int i = 1; i <= n; ++i)
35             if(n % i == 0) reverse(str.begin(), str.begin() + i);
36         cout << str << endl;
37     }
38     return 0;
39 }

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C、Alphabetic Removals

思路：暴力。关联式容器multimap（键---值：一对多）的简单应用。注意：map.erase(迭代器)的结果只是被删元素的迭代器失效，但因为其返回值为void，所以要采用map.erase(it++)的方式删除被删元素的迭代器，此时it已经指向下一个迭代器的位置了。

AC代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <iomanip>
 8 #include <complex>
 9 #include <string>
10 #include <vector>
11 #include <set>
12 #include <map>
13 #include <list>
14 #include <deque>
15 #include <queue>
16 #include <stack>
17 #include <bitset>
18 using namespace std;
19 typedef long long LL;
20 typedef unsigned long long ULL;
21 const int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}}; // 上右下左
22 const int mx[8] = {-1, -2, -2, -1, 1, 2, 2, 1}; // 马可走的八个方向
23 const int my[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
24 const double eps = 1e-6;
25 const double PI = acos(-1.0);
26 const int maxn = 4e5+5;
27 const int inf = 0x3f3f3f3f;
28 
29 int n, k, len;
30 
31 multimap<char, int> mp1;
32 multimap<char, int>::iterator it1;
33 
34 map<int, char> mp2;
35 map<int, char>::iterator it2;
36 
37 char str[maxn];
38 
39 int main() {
40     while(cin >> n >> k >> str) {
41         len = strlen(str);
42         if(k == len) continue;
43         mp1.clear(); mp2.clear();
44         for(int i = 0; i < len; ++i) mp1.insert(make_pair(str[i], i));
45         // 注意：mp1.erase(迭代器)的结果只是被删元素的迭代器失效，但因为其返回值为void，所以要采用erase(it++)的方式删除被删元素的迭代器
46         for(it1 = mp1.begin(); k && (it1 != mp1.end()); --k) mp1.erase(it1++);
47         for(it1 = mp1.begin(); it1 != mp1.end(); ++it1) mp2[it1 -> second] = it1 -> first;
48         for(it2 = mp2.begin(); it2 != mp2.end(); ++it2) printf("%c", it2 -> second);
49         puts("");
50     }
51     return 0;
52 }

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