A. Broken Keyboard
Description
给出一串小写字母字符序列，连续出现两次的字母为坏掉的，按字典序输出所有没有坏掉的字母。
Solution
模拟暴力删除字母，注意相同字母的去重。
1 #include <algorithm>
2 #include <cctype>
3 #include <cmath>
4 #include <cstdio>
5 #include <cstdlib>
6 #include <cstring>
7 #include <iostream>
8 #include <map>
9 #include <numeric>
10 #include <queue>
11 #include <set>
12 #include <stack>
13 #if __cplusplus >= 201103L
14 #include <unordered_map>
15 #include <unordered_set>
16 #endif
17 #include <vector>
18 #define lson rt << 1, l, mid
19 #define rson rt << 1 | 1, mid + 1, r
20 #define LONG_LONG_MAX 9223372036854775807LL
21 #define pblank putchar(‘ ‘)
22 #define ll LL
23 using namespace std;
24 typedef long long ll;
25 typedef long double ld;
26 typedef unsigned long long ull;
27 typedef pair<int, int> P;
28 int n, m, k;
29 const int maxn = 1e5 + 10;
30 template <class T>
31 inline T read()
32 {
33 int f = 1;
34 T ret = 0;
35 char ch = getchar();
36 while (!isdigit(ch))
37 {
38 if (ch == ‘-‘)
39 f = -1;
40 ch = getchar();
41 }
42 while (isdigit(ch))
43 {
44 ret = (ret << 1) + (ret << 3) + ch - ‘0‘;
45 ch = getchar();
46 }
47 ret *= f;
48 return ret;
49 }
50 template <class T>
51 inline void write(T n)
52 {
53 if (n < 0)
54 {
55 putchar(‘-‘);
56 n = -n;
57 }
58 if (n >= 10)
59 {
60 write(n / 10);
61 }
62 putchar(n % 10 + ‘0‘);
63 }
64 template <class T>
65 inline void writeln(const T &n)
66 {
67 write(n);
68 puts("");
69 }
70 int vis[26];
71 int main(int argc, char const *argv[])
72 {
73 #ifndef ONLINE_JUDGE
74 freopen("in.txt", "r", stdin);
75 // freopen("out.txt", "w", stdout);
76 #endif
77 cin >> n;
78 vector<char> res;
79 res.clear();
80 while (n--)
81 {
82 res.clear();
83 string cur;
84 cin >> cur;
85 int sz = cur.size();
86 if (sz == 1)
87 {
88 cout << cur << "\n";
89 continue;
90 }
91 int i = 0;
92 if (cur[0] != cur[1])
93 {
94 i = 1;
95 res.emplace_back(cur[0]);
96 }
97 else
98 i = 2;
99
100 cur += " ";
101 for (; i < sz; i++)
102 if (cur[i] == cur[i + 1])
103 i++;
104 else
105 res.emplace_back(cur[i]);
106 sort(res.begin(), res.end());
107 auto cend = unique(res.begin(), res.end());
108 for (auto it = res.begin(); it != cend; it++)
109 cout << *it;
110 cout << "\n";
111 }
112 return 0;
113 }
B. Binary Palindromes
Description
给出n个01串，串与串之间，串内可以任意交换，求最大能构成多少个回文串。
Solution
考场降智，想到了记录01个数以及奇数长度的串个数，但是最后结论出了问题，忘记考虑偶数个奇数长度的串情况
记录所有串中0出现的次数zero，1出现的次数one
记录奇数长度串的个数odd
对于全是偶数串的情况，如果0和1的个数都是偶数则必定可以构成n个回文串，而0和1为奇数则会由一个串不能构成，答案为n-1
对于含有奇数长度串的情况，如果odd是奇数，那么总的字母数一定是奇数，$odd(zero+one)$,
显然我们先拿出odd个数的0或者1来填充字符串中间，子问题就是构造n个偶数长度的回文串，而剩下的必定能满足one和zero都是偶数，答案为n
对于odd为偶数的情况，$even(one+zero)$,显然按照奇数情况考虑，一定能将zero和one全变为偶数，答案为n
1 #include <algorithm>
2 #include <cctype>
3 #include <cmath>
4 #include <cstdio>
5 #include <cstdlib>
6 #include <cstring>
7 #include <iostream>
8 #include <map>
9 #include <numeric>
10 #include <queue>
11 #include <set>
12 #include <stack>
13 #if __cplusplus >= 201103L
14 #include <unordered_map>
15 #include <unordered_set>
16 #endif
17 #include <vector>
18 #define lson rt << 1, l, mid
19 #define rson rt << 1 | 1, mid + 1, r
20 #define LONG_LONG_MAX 9223372036854775807LL
21 #define pblank putchar(‘ ‘)
22 #define ll LL
23 using namespace std;
24 typedef long long ll;
25 typedef long double ld;
26 typedef unsigned long long ull;
27 typedef pair<int, int> P;
28 int n, m, k;
29 const int maxn = 1e5 + 10;
30 template <class T>
31 inline T read()
32 {
33 int f = 1;
34 T ret = 0;
35 char ch = getchar();
36 while (!isdigit(ch))
37 {
38 if (ch == ‘-‘)
39 f = -1;
40 ch = getchar();
41 }
42 while (isdigit(ch))
43 {
44 ret = (ret << 1) + (ret << 3) + ch - ‘0‘;
45 ch = getchar();
46 }
47 ret *= f;
48 return ret;
49 }
50 template <class T>
51 inline void write(T n)
52 {
53 if (n < 0)
54 {
55 putchar(‘-‘);
56 n = -n;
57 }
58 if (n >= 10)
59 {
60 write(n / 10);
61 }
62 putchar(n % 10 + ‘0‘);
63 }
64 template <class T>
65 inline void writeln(const T &n)
66 {
67 write(n);
68 puts("");
69 }
70 char s[100][100];
71 int len[100];
72 int z[100], o[100];
73 int main(int argc, char const *argv[])
74 {
75 #ifndef ONLINE_JUDGE
76 freopen("in.txt", "r", stdin);
77 // freopen("out.txt", "w", stdout);
78 #endif
79 int t;
80 cin >> t;
81 while (t--)
82 {
83 cin >> n;
84 int zz = 0, oo = 0, o = 0;
85 for (int i = 0; i < n; i++)
86 {
87 string s;
88 cin >> s;
89 len[i] = s.size();
90 if (len[i] & 1)
91 ++o;
92 for (int j = 0; j < len[i]; j++)
93 if (s[j] == ‘0‘)
94 ++zz;
95 else
96 ++oo;
97 }
98 int res = 0;
99 if (o || zz % 2 == 0)
100 res = n;
101 else
102 res = n - 1;
103 writeln(res);
104 }
105 return 0;
106 }
C. Minimize The Integer
Description
给出一个长度为n的数字串，奇偶性不同的相邻两数可以交换，求交换后的最小值。
Solution
奇数和奇数不能交换位置，偶数和偶数可以交换位置
也就是说奇/偶数序列的先后关系是一定的,而我们要做的就是将奇偶序列重新合并，类似归并排序的合并两个子序列
1 #include <algorithm>
2 #include <cctype>
3 #include <cmath>
4 #include <cstdio>
5 #include <cstdlib>
6 #include <cstring>
7 #include <iostream>
8 #include <map>
9 #include <numeric>
10 #include <queue>
11 #include <set>
12 #include <stack>
13 #if __cplusplus >= 201103L
14 #include <unordered_map>
15 #include <unordered_set>
16 #endif
17 #include <vector>
18 #define lson rt << 1, l, mid
19 #define rson rt << 1 | 1, mid + 1, r
20 #define LONG_LONG_MAX 9223372036854775807LL
21 #define pblank putchar(‘ ‘)
22 #define ll LL
23 #define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
24 using namespace std;
25 typedef long long ll;
26 typedef long double ld;
27 typedef unsigned long long ull;
28 typedef pair<int, int> P;
29 int n, m, k;
30 const int maxn = 1e5 + 10;
31 template <class T>
32 inline T read()
33 {
34 int f = 1;
35 T ret = 0;
36 char ch = getchar();
37 while (!isdigit(ch))
38 {
39 if (ch == ‘-‘)
40 f = -1;
41 ch = getchar();
42 }
43 while (isdigit(ch))
44 {
45 ret = (ret << 1) + (ret << 3) + ch - ‘0‘;
46 ch = getchar();
47 }
48 ret *= f;
49 return ret;
50 }
51 template <class T>
52 inline void write(T n)
53 {
54 if (n < 0)
55 {
56 putchar(‘-‘);
57 n = -n;
58 }
59 if (n >= 10)
60 {
61 write(n / 10);
62 }
63 putchar(n % 10 + ‘0‘);
64 }
65 template <class T>
66 inline void writeln(const T &n)
67 {
68 write(n);
69 puts("");
70 }
71 void solve(const string &s)
72 {
73 string res, odd, even;
74 res.clear(), odd.clear(), even.clear();
75 int len = s.size();
76 for (int i = 0; i < len; i++)
77 if (s[i] & 1)
78 odd += s[i];
79 else
80 even += s[i];
81 int sz1 = odd.size(), sz2 = even.size();
82 int i = 0, j = 0;
83 while (i < sz1 && j < sz2)
84 {
85 if (odd[i] < even[j])
86 cout << odd[i++];
87 else
88 cout << even[j++];
89 }
90 while (i < sz1)
91 cout << odd[i++];
92 while (j < sz2)
93 cout << even[j++];
94 cout << "\n";
95 }
96 int main(int argc, char const *argv[])
97 {
98 #ifndef ONLINE_JUDGE
99 freopen("in.txt", "r", stdin);
100 // freopen("out.txt", "w", stdout);
101 #endif
102 fast;
103 int t;
104 cin >> t;
105 while (t--)
106 {
107 string s;
108 cin >> s;
109 solve(s);
110 }
111 return 0;
112 }
D. Salary Changing
Description
给出n个闭区间，一个上限s，在每个区间里取一个数使得这些数的和不大于上限s且其中位数最大。
Solution
二分答案中位数x
先按照左端点排序，将区间划分成三种情况，$l:$严格小于x，$r:$严格大于x，$mid:$包含x
对于第一,二种区间，直接累加区间左值,记录对应种类区间取数的次数$l,r$
如果l或者大于n/2，则中位数不可能是x，fail
我们要的结果应当是$l=r=n/2 \ and \ sum \leq s$
接下来考虑第三种区间
从左往右遍历，满足区间排序的情况下，如果$l \lt n/2，sum+=now.l$，否则如果$r \lt n/2$，$sum+=x$
这样的取法满足sum最小且中位数为check值
1 #include <algorithm>
2 #include <cctype>
3 #include <cmath>
4 #include <cstdio>
5 #include <cstdlib>
6 #include <cstring>
7 #include <iostream>
8 #include <map>
9 #include <numeric>
10 #include <queue>
11 #include <set>
12 #include <stack>
13 #if __cplusplus >= 201103L
14 #include <unordered_map>
15 #include <unordered_set>
16 #endif
17 #include <vector>
18 #define lson rt << 1, l, mid
19 #define rson rt << 1 | 1, mid + 1, r
20 #define LONG_LONG_MAX 9223372036854775807LL
21 #define pblank putchar(‘ ‘)
22 #define ll LL
23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
24 using namespace std;
25 typedef long long ll;
26 typedef long double ld;
27 typedef unsigned long long ull;
28 typedef pair<ll, ll> P;
29 int n, m, k;
30 const int maxn = 2e5 + 10;
31 template <class T>
32 inline T read()
33 {
34 int f = 1;
35 T ret = 0;
36 char ch = getchar();
37 while (!isdigit(ch))
38 {
39 if (ch == ‘-‘)
40 f = -1;
41 ch = getchar();
42 }
43 while (isdigit(ch))
44 {
45 ret = (ret << 1) + (ret << 3) + ch - ‘0‘;
46 ch = getchar();
47 }
48 ret *= f;
49 return ret;
50 }
51 template <class T>
52 inline void write(T n)
53 {
54 if (n < 0)
55 {
56 putchar(‘-‘);
57 n = -n;
58 }
59 if (n >= 10)
60 {
61 write(n / 10);
62 }
63 putchar(n % 10 + ‘0‘);
64 }
65 template <class T>
66 inline void writeln(const T &n)
67 {
68 write(n);
69 puts("");
70 }
71 vector<P> seg;
72 ll s;
73 int dep;
74 int vis[maxn];
75 inline int check(ll cur)
76 {
77 int l = 0, r = 0, mid = 0;
78 ll sum = 0;
79 memset(vis, 0, sizeof(int) * (n + 1));
80 for (int i = 0; i < n; i++)
81 {
82 if (seg[i].first <= cur && seg[i].second >= cur)
83 ++mid;
84 else if (seg[i].second < cur)
85 {
86 vis[i] = 1;
87 ++l;
88 sum += seg[i].first;
89 }
90 else if (seg[i].first > cur)
91 {
92 ++r;
93 vis[i] = 1;
94 sum += seg[i].first;
95 }
96 }
97 if (l > dep || r > dep)
98 return 0;
99 for (int i = 0; i < n; i++)
100 if (l < dep && !vis[i])
101 ++l, sum += seg[i].first;
102 else if (r < dep && !vis[i])
103 ++r, sum += cur;
104 sum += cur;
105 return sum <= s;
106 }
107 int main(int argc, char const *argv[])
108 {
109 #ifndef ONLINE_JUDGE
110 freopen("in.txt", "r", stdin);
111 // freopen("out.txt", "w", stdout);
112 #endif
113 int t = read<int>();
114 while (t--)
115 {
116 seg.clear();
117 n = read<int>();
118 s = read<ll>();
119 dep = n >> 1;
120 for (int i = 1; i <= n; i++)
121 {
122 ll x = read<ll>(), y = read<ll>();
123 seg.emplace_back(x, y);
124 }
125 sort(seg.begin(), seg.end());
126 ll l = seg[dep].first, r = s;
127 ll res = -1;
128 while (l <= r)
129 {
130 ll mid = l + r >> 1;
131 if (check(mid))
132 {
133 l = mid + 1;
134 res = mid;
135 }
136 else
137 r = mid - 1;
138 }
139 writeln(res);
140 }
141 return 0;
142 }
熬夜掉分场
原文地址：https://www.cnblogs.com/mooleetzi/p/11739173.html
时间： 2024-11-05 05:33:47