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 << " "; 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 << " "; 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 << " "; 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