A Raining
void solve() { int a[4], b[4] = {0}; for (int i = 0; i <= 3; i++) {cin >> a[i]; } int x; cin >> x; for (int i = 0; i <= 3; i++) { if(a[i] < x) b[i] = x - a[i]; } for (int i = 0; i <= 3; i++) {cout << b[i] << ' '; } }
B Kissing
- 化简
s = int(input()) mod = 998244353 print(s % mod * s % mod）
C Missing
题目
代码
- 赛后
#include <bits/stdc++.h> using namespace std; int main() { string s; cin >> s; int n; cin >> n; vector<string> a[11]; auto f = [&](string A){ int res = 0; for (int i = 0; i < s.size(); i++) res += (A[i] == s[i]); if(A.size() != s.size()) res = 0; return res; }; for (int i = 0; i < n; i++) { string t; cin >> t; a[f(t)].push_back(t); } for (int i = 10; i >= 0; i--) { if(a[i].size()) { sort(a[i].begin(), a[i].end()); for (auto &x: a[i]) cout << x << endl; break; } } return 0; }
- 赛时，没想到用相似度索引。
string s[N]; double b[N]; map<string, double> mp; void solve() { string a; cin >> a; int n; cin >> n; for (int i = 0; i < n; i++) cin >> s[i]; for (int i = 0; i < n; i++) { int cnt = 0; if(s[i].sz != a.sz) continue; for (int j = 0; j < s[i].sz; j++) { if(s[i][j] == a[j]) cnt++; } b[i] = cnt * 1.0 / s[i].sz; mp[s[i]] = b[i]; } sort(s, s + n, [=](string i, string j){ if(mp[i] == mp[j]) return i < j; return mp[i] > mp[j]; }); cout << s[0] << endl; for (int i = 1; i < n; i++) { if(mp[s[i]] != mp[s[0]]) break; cout << s[i] << endl; } }
D Breezing
题目
代码
int n; int a[N]; int dp[N][2]; void solve() { cin >> n; for (int i = 1; i <= n; i++) cin >> a[i]; dp[1][0] = dp[1][1] = 0; for (int i = 2; i <= n; i++) { // if(dp[i - 1][0] > dp[i - 1][1] + a[i - 1] - 1) dp[i][0] = dp[i - 1][0]; // else dp[i][0] = dp[i - 1][1] + a[i - 1] - 1; // if(dp[i - 1][0] + a[i] - 1 > dp[i - 1][1] + abs(a[i - 1] - a[i])) dp[i][1] = dp[i - 1][0] + a[i] - 1; // else dp[i][1] = dp[i - 1][1] + abs(a[i - 1] - a[i]); dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] + a[i - 1] - 1); dp[i][1] = max(dp[i - 1][0] + a[i] - 1, dp[i - 1][1] + abs(a[i] - a[i - 1])); } cout << max(dp[n][1], dp[n][0]) << endl; }
E Calling
题目
代码
- 先算得3、4、5、6装下所需要得大正方形个数res。
- 在算得剩余空位能装2的个数res2，如果 a[2] > res2，换大正方形装2。
- 如此算1。
#include <iostream> using namespace std; int main() { int t; cin >> t; while(t--) { int a[7] = {0}; int n; cin >> n; for (int i = 1; i <= 6; i++) cin >> a[i]; int res = 0; for (int i = 4; i <= 6; i++) res += a[i]; res += (a[3] + 3) / 4; int res2 = a[4] * 5; if(a[3] % 4 == 1) res2 += 5; else if(a[3] % 4 == 2) res2 += 3; else if(a[3] % 4 == 3) res2 += 1; if(a[2] > res2) res += (a[2] - res2 + 8) / 9; int res1 = res * 36; for (int i = 2; i <= 6; i++) res1 -= a[i] * i * i; if(a[1] > res1) res += (a[1] - res1 + 35) / 36; puts(n >= res ? "Yes" : "No"); } return 0; }
