AlCu牛客小白月赛55 A-E
https://ac.nowcoder.com/acm/contest/38630
F待补
解析啥的睡醒再补，先放个代码
A - 至至子的等差中项
#include <bits/stdc++.h>using namespace std;int main () {int a, b;cin >> a >> b;cout << 2*b-a;
}
B - 至至子的按位与
#include <bits/stdc++.h>
#define int long longusing namespace std;
int a, b;void solve () {cin >> a >> b;int maxn = (1ll << 63) - 1ll;cout << maxn - (a ^ b) << endl;
}signed main () {solve ();
}
C - 至至子的斐波那契
#include <bits/stdc++.h>
#define int long longusing namespace std;
vector <int> v;void pre () {v.push_back (0), v.push_back (1), v.push_back (1);int x = 0;for (int i = 3; x <= 1e18; i++) {x = v[i-1] + v[i-2];v.push_back (x);}//for (auto i : v) cout << i << ", ";
}void solve () {int n, i;cin >> n;for (i = 1; ; i++) {if (v[i] >= n) break;}if (v[i] - n >= n - v[i-1]) cout << i-1;else cout << i;cout << endl;
}signed main () {pre ();int t;cin >> t;while (t --) {solve ();}
}
D - 至至子的鸿门宴
#include <bits/stdc++.h>
#define int long longusing namespace std;
const int N = 1e6 + 5;
int n, a[N];signed main () {cin >> n;int sum = 0;for (int i = 1; i <= n; i++) {cin >> a[i];sum += a[i] - i;}// b[1] = a[1];// for (int i = 1; i <= n; i++) b[i] --, sum += b[i];//cout << sum << endl;if (sum % 2 == 0) puts ("SSZ");else puts ("ZZZ");}
E - 至至子的长链剖分
#include <bits/stdc++.h>
#define int long longusing namespace std;
typedef pair<int, int> pii;
const int N = 2e5 + 5;
int n;void solve () {cin >> n;vector <int> cnt[n+1]; //pre cntint maxn = 0, x;for (int i = 1; i <= n; i++) {cin >> x;cnt[x].push_back (i);maxn = max (maxn, x);}// for (int i = 0; i <= n; i++) {// if (cnt[i].size() == 0) continue;// cout << "i= " << i << ", ";// for (auto j : cnt[i])// cout << j << ' ';// cout << endl;// }if (n == 1) {if (x) cout << -1 << endl;else cout << 1 << endl;return ;}if (cnt[maxn].size() > 1 || cnt[0].size() == 0) {cout << -1 << endl;return ;}vector <pii> ans;for (int i = 1; i <= maxn; i++) {if (cnt[i].size() == 0 || cnt[i].size() > cnt[i-1].size()) {cout << "-1\n";return ;}int cnt1 = cnt[i-1].size(), cnt2 = cnt[i].size();for (int j = 0; j < cnt2; j++) ans.push_back ({cnt[i-1][j], cnt[i][j]});for (int j = cnt2; j < cnt1; j++) ans.push_back ({cnt[i-1][j], cnt[i][0]});}cout << cnt[maxn][0] << endl;for (auto i : ans) cout << i.first << ' ' << i.second << endl;}signed main () {int t;cin >> t;while (t --) {solve ();}
}//h为几就在第几+1层(从下往上)
//相邻数字连一条边
//并且上层的个数一定要小于等于下层的//之前是把所有的都连到一点上所以就错了
//上一层与下一层尽量相连，然后多的在统一连到一点
F - 至至子的公司排队
别急
