2017 ACM/ICPC Asia Regional Qingdao Online 记录

题目链接  Qingdao

Problem C

AC自动机还不会，暂时暴力水过。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)

const int N = 1e5 + 10;

string s[N];
int T;
int n;
int ans;

int main(){

	std::ios::sync_with_stdio(false);

	cin >> T;
	while (T--){
		cin >> n;
		int id;
		int maxv = 0;
		for(int i = 1; i <= n; i++) {
			cin >> s[i];
			if (s[i].size() > maxv) {
				maxv = s[i].size();
				id = i;
			}
		}
		ans = 1;
		rep(i, 1, n) if (s[id].find(s[i]) == -1){ ans = 0; break;}
		if (ans) cout << s[id] << endl;
		else cout << "No" << endl;
	}

	return 0;
}

Problem J

考虑直接用队列保存待判断的元素（出队or not）

然后直接用链表模拟就可以了。

为什么比赛的时候我不会做呢

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second

typedef long long LL;

const int N = 1e5 + 10;

struct node{
	int x, l, r;
} a[N];

int T;
int n, ans;
queue <int> q;

int main(){

	scanf("%d", &T);
	while (T--){
		scanf("%d", &n);
		rep(i, 1, n){
			a[i].l = i - 1;
			scanf("%d", &a[i].x);
			a[i].r = i + 1;
		}

		a[0].r = 1;
		a[n + 1].l = n;
		a[0].x = 0;
		a[n + 1].x = 1e8;

		while (!q.empty()) q.pop();

		rep(i, 1, n) q.push(i);
		while (!q.empty()){
			int now = q.front(); q.pop();
			int suc = a[now].r;
			int pre = a[now].l;
			if (a[now].x > a[suc].x){
				q.push(pre);
				a[pre].r = a[suc].r;
				a[a[suc].r].l = pre;
				a[suc].l = pre;
			}
		}

		ans = 0;
		int now = a[0].r;
		while (now <= n){
			++ans;
			now = a[now].r;
		}

		printf("%d\n", ans);
		now = a[0].r;
		while (now <= n){
			printf("%d ", a[now].x);
			now = a[now].r;
		}

		putchar(10);
	}

	return 0;
}

Problem K

签到

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second

typedef long long LL;

LL a[1001000], b[1001000];
LL n;
int T;

int main(){

	for (LL i = 1; i <= 1000000; ++i) a[i] = i * i * i;
	rep(i, 1, 999999) b[i] = a[i + 1] - a[i];

	scanf("%d", &T);
	while (T--){
		scanf("%lld", &n);
		bool fl = false;
		rep(i, 1, 999999) if (b[i] == n){
		       fl = true;
	       		 break;
		}

		if (fl) puts("YES"); else puts("NO");
	}		

	return 0;
}