总结：感觉这个educational场好难啊（蒟蒻视角），又被虐了，唉

A. Dungeon

每一枪会造成1点伤害对一个单位，但是当开7的倍数枪时会造成3点伤害
每7次一个轮回，一个完整的轮回共造成9点伤害，如果最后一起死，那个最小值至少需要承受每个轮回的1点伤害

#include <bits/stdc++.h>
#define lowbit(x) ((x) & (-x))
using namespace std;typedef long long ll;int main() {#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);
#endifint t;cin >> t;while (t--) {int a, b, c;cin >> a >> b >> c;int min1 = min(a, min(b, c));ll sum = a + b + c;if (sum % 9 == 0 && min1 >= sum / 9) cout << "YES" << endl;else cout << "NO" << endl;}return 0;
}

B - Find The Array

哇，这个题可真是够难，想不出来
最后经大神提醒，每个数直接取他的二进制最高位就可以，既保证了倍数，又保证了总和不会超过原来的一半

#include <bits/stdc++.h>
#define lowbit(x) ((x) & (-x))
using namespace std;typedef long long ll;
const int N = 100;int a[N], b[N];int solve(int x) {int cnt = 0;while (x) {cnt++;x >>= 1;}  return 1 << (cnt - 1);
}int main() {#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);
#endifint t;cin >> t;while (t--) {int n;cin >> n;for (int i = 1; i <= n; i++) {cin >> a[i];b[i] = solve(a[i]);}for (int i = 1; i <= n; i++) {if (i != 1) cout << " ";cout << b[i];}cout << endl;}return 0;
}

C - Busy Robot

模拟题

最后的时间时是正无穷
一个命令若是要执行成功，那么必须在下一个命令要求的时间之前（包括端点）到达这个命令的终点
一个命令执行时，会忽略在此期间发过来的命令（虽然被忽略，但是也可能成功执行，符合上一个条件即可）
一个被忽略的命令若想成功执行，则至少需要包含在正在行进的路径中

#include <bits/stdc++.h>
#define lowbit(x) ((x) & (-x))
using namespace std;typedef long long ll;
const int N = 1e5 + 10;
const int inf = 0x3f3f3f3f;
#define int long long
struct node {ll t, pos;
}e[N];
int book[N];signed main() {#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);
#endifint t;scanf("%lld", &t);while (t--) {int n;scanf("%lld", &n);memset(book, 0, sizeof book);for (int i = 1; i <= n; i++) scanf("%lld%lld", &e[i].t, &e[i].pos);e[n + 1].t = 1e18; int cnt = 0, st = 0;for (int i = 1; i <= n; i++) {if (!book[i]) {book[i] = 1;if ((e[i + 1].t - e[i].t >= abs(e[i].pos - st))) cnt++;  int ed = e[i].t + abs(e[i].pos - st);int j = i + 1;while (j <= n && e[j].t < ed) {book[j] = 1;int flag = 0;if (e[j].pos >= st && e[j].pos <= e[i].pos) flag = 1;if (e[j].pos >= e[i].pos && e[j].pos <= st) flag = 1;if (flag && (e[j].t - e[i].t) <= abs(e[j].pos - st) && (e[i].t + abs(e[j].pos - st) <= e[j + 1].t)) cnt++;j++;}st = e[i].pos;}}printf("%lld\n", cnt);} // return 0;
}

D - Pairs

这位大佬的思路 https://blog.csdn.net/ANTFANAAA/article/details/111351485

也是模拟题

有些点只能作为最小值出现
有些点只能作为最大值出现
有些点既能作为最小值，也能作为最大值
那么怎么求重合的这部分呢？那就是ans1 + ans2 - n
最终是求最小值包含的数的所有情况，那就是重合部分 + 初始值（也就是1）

#include <bits/stdc++.h>
#define lowbit(x) ((x) & (-x))
using namespace std;typedef long long ll;
const int inf = 0x3f3f3f3f;
// #define int long long
const int N = 1e6 + 10;
int book[N], a[N];int main() {#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);
#endifint t;scanf("%d", &t);while (t--) {int n;scanf("%d", &n);set<int> v1, v2;for (int i = 1; i <= 2 * n; i++) book[i] = 0;for (int i = 1; i <= n; i++) {scanf("%d", &a[i]);book[a[i]] = 1;}for (int i = 1; i <= 2 * n; i++) {if (!book[i]) {v1.insert(i);v2.insert(i);}}int ans1 = 0, ans2 = 0;for (int i = 1; i <= n; i++) {auto t1 = v1.lower_bound(a[i]);auto t2 = v2.lower_bound(a[i]);if (t1 != v1.end()) {ans1++;v1.erase(t1);}if (t2 != v2.begin()) {ans2++;t2--;v2.erase(t2);}}printf("%d\n", (ans1 + ans2 - n) + 1);}return 0;
}

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Educational Codeforces Round 100 (Rated for Div. 2)补题记录

总结：感觉这个educational场好难啊（蒟蒻视角），又被虐了，唉

A. Dungeon

B - Find The Array

C - Busy Robot

D - Pairs

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