Codeforces 903 绝对值1e19longdouble算贡献 汉明距离交换两项是否可行
A
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7; const int gakki = 5 + 2 + 1 + 19880611 + 1e9; const int N = (1 << 22) + 5; int main() { int T; cin >> T; while (T--) { int flag=0; int n; cin >> n; for (int i = 0; i <= n / 3; i++) { if ((n - 3 * i) % 7 == 0) { cout << "YES" << endl; flag=1; break; } } if(!flag) { cout<<"NO"<<endl; } } return 0; }
B
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7; const int gakki = 5 + 2 + 1 + 19880611 + 1e9; const int N = (1 << 22) + 5; string ans[100005]; int cnt = 0; int main() { ios_base::sync_with_stdio(0); cin.tie(0); int h1, a1, c1; int h2, a2; cin >> h1 >> a1 >> c1; cin >> h2 >> a2; while (h2 > 0) { if (h2 - a1 <= 0) { ans[++cnt] = "STRIKE"; h2 -= a1; continue; } if (h1 - a2 <= 0) { ans[++cnt] = "HEAL"; h1 += c1-a2; continue; } ans[++cnt] = "STRIKE"; h1 -= a2; h2 -= a1; } cout << cnt << endl; for (int i = 1; i <= cnt; i++) { cout << ans[i] << endl; } return 0; }
C
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7; const int gakki = 5 + 2 + 1 + 19880611 + 1e9; const int N = 5e3 + 5; int vis[N]; int num[N]; int visit[N]; int main() { ios_base::sync_with_stdio(0); cin.tie(0); int n; cin >> n; int maxn = -1; for (int i = 1; i <= n; i++) { cin >> num[i]; maxn = max(maxn, num[i]); } sort(num + 1, num + 1 + n); for (int i = 1; i <= n; i++) { if (visit[i]) { continue; } int cur = num[i]; int pre = i; vis[pre] = 1; visit[pre] = 1; for (int j = i + 1; j <= n; j++) { if (num[j] > cur && !visit[j]) { visit[j] = 1; vis[pre] = 0; cur = num[j]; pre = j; vis[pre] = 1; } } } int anser = 0; for (int i = 1; i <= n; i++) { if (vis[i]) { anser++; } } cout << anser << endl; return 0; }
D
会爆long long 要用long double
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) #define pb push_back using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7; const int gakki = 5 + 2 + 1 + 19880611 + 1e9; const int N = 2e5 + 5; map<double, double> mp; long double num[N]; long double now; long double sheng; int main() { ios_base::sync_with_stdio(0); cin.tie(0); long double sum = 0; long double anser = 0; int n; cin >> n; for (int i = 1; i <= n; i++) { cin >> num[i]; sum += num[i]; mp[num[i]]++; } for (int i = 1; i < n; i++) { sum -= num[i - 1]; if (i != 1) { mp[num[i - 1]]--; } sheng = n - i + 1 - mp[num[i] - 1] - mp[num[i] + 1]; anser += sum - mp[num[i] - 1] * (num[i] - 1) - mp[num[i] + 1] * (num[i] + 1) - sheng * num[i]; } cout << fixed << setprecision(0) << anser << endl; return 0; }
JAVA版
import java.io.OutputStream; import java.io.PrintWriter; import java.math.BigInteger; import java.util.*;; public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); OutputStream outputStream = System.out; PrintWriter out = new PrintWriter(outputStream); SortedMap<Long, Long>ma = new TreeMap<>() ; BigInteger result = new BigInteger("0"); BigInteger currentSum = new BigInteger ("0"); int n = in.nextInt(); for (int i = 0; i < n; i++) { long x = in.nextInt(); long count = i - (ma.getOrDefault(x - 1,0L) + ma.getOrDefault(x,0L) + ma.getOrDefault(x + 1,0L)); long toRemove = (x - 1) * ma.getOrDefault(x - 1,0L) + (x)* ma.getOrDefault(x,0L) + (x + 1) * ma.getOrDefault(x + 1,0L); BigInteger valueY = currentSum.subtract(new BigInteger(String.valueOf(toRemove))); result = result.add(new BigInteger(String.valueOf((x * count)))).subtract(valueY); currentSum = currentSum.add(new BigInteger(String.valueOf(x))); ma.put(x, ma.getOrDefault(x,0L)+1); } out.println(result.toString()); in.close(); out.close(); } }
E
题意:
给你K个长度为N的字符串 每个字符串可以交换任意两项一次 问你这N个字符串能不能最后都变为一个字符串
解:
以第一个字符串为主串 N枚举每个其他串 测出与其他串的汉明距离
然后K2枚举交换的位置 再N枚举其他串 看交换后是否满足
满足有两种情况 1.第一个字符串有两个相同字母 交换后的汉明距离为0 2.交换后的汉明距离为2
当当前交换的两项可以满足所有其他串时 交换这两项后的字符串就是目标串
/*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; const int dir[8][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}, {1, 1}, {1, -1}, { -1, -1}, { -1, 1}}; const int mod = 1e9 + 7, gakki = 5 + 2 + 1 + 19880611 + 1e9; const int MAXN = 1e5 + 5, MAXM = 1e5 + 5, N = 1e4 + 5; const int MAXQ = 100010; int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], tot = 1; inline void addedge(int u, int v) { to[++tot] = v; nxt[tot] = Head[u]; Head[u] = tot; } inline void read(int &v) { v = 0; char c = 0; int p = 1; while (c < '0' || c > '9') { if (c == '-') { p = -1; } c = getchar(); } while (c >= '0' && c <= '9') { v = (v << 3) + (v << 1) + c - '0'; c = getchar(); } v *= p; } string st[3005]; int num[2][30]; int hmdis[3000]; int n, k; int flag = 0; int main() { ios_base::sync_with_stdio(0); cin.tie(0); cin >> k >> n; for (int i = 1; i <= k; i++) { cin >> st[i]; } for (int i = 0; i < n; i++) { num[0][st[1][i] - 'a']++; if (num[0][st[1][i] - 'a'] > 1) { flag = 1; } } for (int i = 2; i <= k; i++) { mem(num[1], 0); for (int j = 0; j < n; j++) { num[1][st[i][j] - 'a']++; if (st[1][j] != st[i][j]) { hmdis[i]++; } } for (int j = 0; j <= 25; j++) { if (num[1][j] != num[0][j]) { cout << -1 << endl; return 0; } } } for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { int now = 0; for (int w = 1; w <= k; w++) { int cnt = hmdis[w]; if (st[1][i] == st[w][i]) { cnt++; } if (st[1][j] == st[w][j]) { cnt++; } if (st[1][i] == st[w][j]) { cnt--; } if (st[1][j] == st[w][i]) { cnt--; } if (cnt == 2 || (cnt == 0 && flag)) { now++; } else { break; } } if (now == k) { swap(st[1][i], st[1][j]); cout << st[1] << endl; return 0; } } } cout << -1 << endl; return 0; }
