Codeforces Round #550 (Div. 3)
A、Diverse Strings
思路：简单判断一下给定的字符串是否刚好为一个顺子即可。
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, len, now, cnt[30]; 30 char str[105]; 31 bool flag, first; 32 map<int, int> mp; 33 34 int main() { 35 while(cin >> n) { 36 while(n--) { 37 cin >> str; 38 memset(cnt, 0, sizeof(cnt)); 39 len = strlen(str); 40 flag = first = false; 41 mp.clear(); 42 for(int i = 0; i < len; ++i) ++mp[str[i] - 'a']; 43 for(map<int, int>::iterator it = mp.begin(); it != mp.end(); ++it) { 44 if(it -> second > 1) {flag = true; break;} 45 if(!first) first = true, now = it -> first; 46 else if(it -> first - now != 1) {flag = true; break;} 47 else now = it -> first; 48 } 49 puts(!flag ? "Yes" : "No"); 50 } 51 } 52 return 0; 53 }
B、Parity Alternated Deletions
思路：贪心。能构造出来的奇偶交替序列的长度一定为2*min(num1,num2)+1，剩下的元素一定是多出来的那几个最小（都是奇数或者都是偶数）的元素之和。
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 = 2005; 27 const int inf = 0x3f3f3f3f; 28 29 int n, x, ans, num1, num2, sum; 30 vector<int> odd, even; 31 32 int main() { 33 while(cin >> n) { 34 odd.clear(), even.clear(); 35 for(int i = 1; i <= n; ++i) { 36 cin >> x; 37 if(x & 1) odd.push_back(x); 38 else even.push_back(x); 39 } 40 sort(odd.begin(), odd.end()); 41 sort(even.begin(), even.end()); 42 num1 = odd.size(); 43 num2 = even.size(); 44 if(abs(num1 - num2) < 2) cout << 0 << endl; 45 else { 46 ans = inf; 47 if(num1 > num2) { 48 num1 -= num2 + 1, sum = 0; 49 for(int i = 0; i < num1; ++i) sum += odd[i]; 50 ans = min(ans, sum); 51 }else { 52 num2 -= num1 + 1, sum = 0; 53 for(int i = 0; i < num2; ++i) sum += even[i]; 54 ans = min(ans, sum); 55 } 56 cout << ans << endl; 57 } 58 } 59 return 0; 60 }
C、Two Shuffled Sequences
思路：简单排序。显然，给定序列中每个数字出现的次数不超过2次；然后简单搞一下即可。
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 30 int n, x, pos, num; 31 vector<int> vec, now; 32 map<int, int> mp; 33 bool flag; 34 35 int main() { 36 while(cin >> n) { 37 vec.clear(), mp.clear(); flag = false; now.clear(); 38 for(int i = 0; i < n; ++i) {cin >> x; vec.push_back(x); ++mp[x];} 39 for(map<int, int>::iterator it = mp.begin(); it != mp.end(); ++it) { 40 if(it -> second > 2) {flag = true; break;} 41 } 42 if(flag) {cout << "NO" << endl; continue;} 43 cout << "YES" << endl; 44 sort(vec.begin(), vec.end()); 45 now.insert(now.begin(), vec.begin(), vec.end()); // 拷贝一份 46 pos = unique(now.begin(), now.end()) - now.begin(); // 去重 指向第一个重复元素的位置 47 cout << pos << endl; 48 for(int i = 0; i < pos; ++i) cout << now[i] << " \n"[i == pos - 1]; 49 cout << n - pos << endl; 50 num = 0; 51 for(map<int, int>::reverse_iterator it = mp.rbegin(); it != mp.rend(); ++it) { 52 if(it -> second == 2) cout << it -> first << " \n"[++num == n - pos]; 53 } 54 if(n - pos == 0) cout << endl; 55 } 56 return 0; 57 }
D、Equalize Them All
思路：贪心。按给定的2种操作使用最少的次数将数组变成n个相同的元素，题目已保证操作后一定能得到n个相同的元素。显然，我们需要将其他元素都变成出现次数最多（cnt个）的数字，这样就要操作n-cnt次，简单搞一下即可。
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 = 2e5+5; 27 const int inf = 0x3f3f3f3f; 28 29 int n, val, cnt, pos, a[maxn]; 30 map<int, int> mp; 31 32 int main() { 33 while(cin >> n) { 34 mp.clear(); 35 for(int i = 0; i < n; ++i) { 36 cin >> a[i]; 37 ++mp[a[i]]; 38 } 39 cnt = 1, val = a[0]; 40 for(map<int, int>::iterator it = mp.begin(); it != mp.end(); ++it) { 41 if(it -> second > cnt) { 42 cnt = it -> second, val = it -> first; 43 } 44 } 45 cout << n - cnt << endl; 46 pos = find(a, a + n, val) - a; // 找到第一个val出现的位置pos 47 for(int i = pos + 1; i < n; ++i) { 48 if(a[i] == val) continue; 49 cout << (1 + (a[i] > val)) << ' ' << i + 1 << ' ' << i << endl; 50 } 51 for(int i = pos - 1; ~i; --i) { 52 if(a[i] == val) continue; 53 cout << (1 + (a[i] > val)) << ' ' << i + 1 << ' ' << i + 2 << endl; 54 } 55 } 56 return 0; 57 }
