QBXT模拟赛2

QBXT模拟赛2

总结

期望得分：\(100 + 40 + 0 = 140\)

实际得分：\(0 + 0 + 0 = 0\)

鬼知道为什么我代码没有交上。。自测\(10 + 50 + 0\)……这是心态爆炸的一场考试

\(T1\)敲了\(3\)个小时，写了\(9kb\)，连样例都过不了，改了改过了样例迅速离开

\(T2\)只会\(40\)，但因为一个点的\(a[i]\)很大，于是水过去了，变成了\(50\)

\(T3\)连看都没看，\(0\)分滚粗

我果是个彩笔

思路&&代码

T1

模拟题， 一看这长长的题面，就知道是个简单题（fp嘛我做了三个小时不过样例能是简单题？）

其实我觉得最难的不是别的，就是输入。。。这么一大堆字符串，还有\(5\)个位置，还是两个队，如何做到最简？

其实这些可以化作一个整体，我们很容易想到要用结构体存储，但复杂的hapigou就是：结构体数组怎么开？

我们可以开一个三维的\(a\)数组，存储两个队所有位置所有人的信息，\(a[0]\)就表示\(A\)队，\(a[1]\)就表示\(B\)队，\(a[x][1-5]\)就表示\(x\)队五个位置的信息，\(a[x][y][cnt]\)就表示\(x\)队第\(y\)个位置的第\(cnt\)名选手，这样不仅好理解，还方便了之后我们要用到的的排序

再者，肯定要从实力由高到低的选，每次某个位置换的肯定是这个位置剩余的实力最强的人，所以我们就可以直接按照实力把每个位置的人排序，如果实力相同则编号较小的在前

之后我们就要想怎么来选人了，要想让场上所有人的综合实力最大，我们要保证换下来的人与在场的人实力的差值最小，所以就可以在五个位置中找差值最小的位置来换，这样这道题就完美解决了

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int A = 5e5 + 11;
const int B = 1e6 + 11;
const int INF = 1000;

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
	return x * f;
}

struct node { char name[16]; int id, val; } a[2][6][A];
int N, M, T, P, Q;
int cnt[2][6], change[2][6];

bool cmp(node a, node b) {
	return a.val == b.val ? a.id < b.id : a.val > b.val;
}

inline int check(char s[4]) {
	if(s[0] == 'p' && s[1] == 'g') return 1;
	if(s[0] == 's' && s[1] == 'g') return 2;
	if(s[0] == 's' && s[1] == 'f') return 3;
	if(s[0] == 'p' && s[1] == 'f') return 4;
	if(s[0] == 'c') return 5;
}

inline void add(int dui, char name[16], int pos, int id, int val) {
	a[dui][pos][++cnt[dui][pos]].id = id;
	strcpy(a[dui][pos][cnt[dui][pos]].name, name);
	a[dui][pos][cnt[dui][pos]].val = val;
}

int main() {
	N = read(), M = read(), T = read(), P = read(), Q = read();
	char s[4], name[16]; int idd, val, dui;
	for(int i = 1; i <= N + M; i++) {
		dui = (i <= N ? 0 : 1);
		scanf("%s %d %s %d", name, &idd, s, &val);
		add(dui, name, check(s), idd, val);
	}
	for(int i = 1; i <= 5; i++) sort(a[0][i] + 1, a[0][i] + cnt[0][i] + 1, cmp), sort(a[1][i] + 1, a[1][i] + cnt[1][i] + 1, cmp);
	for(int i = 1; i <= 5; i++) change[1][i] = 1, change[0][i] = 1;
	int suma = 1, sumb = 1, team;
	while(P * suma < T || Q * sumb < T) {
		if(P * suma <= Q * sumb) team = 0, suma++; else team = 1, sumb++;
		int minn = 10000000, po; node be, aft;
		for(int i = 1; i <= 5; i++) {
			int k = change[team][i], c = a[team][i][k].val - a[team][i][k + 1].val;
			if(k == cnt[team][i]) continue;
			if(c < minn || (c == minn && a[team][i][k + 1].id < aft.id)) minn = c, po = i, be = a[team][i][k], aft = a[team][i][k + 1];
		}
		change[team][po]++;
		char dui = (team == 0 ? 'A' : 'B');
		printf("Substitution by %c,No.%d %s is coming up to change No.%d %s.\n", dui, aft.id, aft.name, be.id, be.name);
	}
	return 0;
}

T2

这道题到现在我还是只会五十分，很容易想到是一个贪心的思路，我们直接用两个指针表示从左边右边拿，就能拿五十了

满分的也在下面，还是不会。。菜死了

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int A = 1e5 + 11;
const int B = 1e6 + 11;
const int INF = 0x3f3f3f3f;

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
	return x * f;
}

int n, T, x[A], a[A];

namespace sub {
	int maxn = -INF;
	inline void Main() {
		for(int i = 1; i <= n; i++) {
			int now = 0, ans = 0, l = i - 1, r = i + 1, one;
			while(now <= T) {
				if(l < 1 && r > n) break;
				else if(l >= 1 && r <= n && x[i] - x[l] <= x[r] - x[i]) {
					one = min((T - now) / ((x[i] - x[l]) * 2), a[l]);
					now += (x[i] - x[l]) * 2 * one, ans += one;
					if(one == a[l]) l--; else break;
				}
				else if(l >= 1 && r <= n && x[i] - x[l] > x[r] - x[i]) {
					one = min((T - now) / ((x[r] - x[i]) * 2), a[r]);
					now += (x[r] - x[i]) * 2 * one, ans += one;
					if(one == a[r]) r++; else break;
				}
				else if(l < 1 && r <= n) {
					one = min((T - now) / ((x[r] - x[i]) * 2), a[r]);
					now += (x[r] - x[i]) * 2 * one, ans += one;
					if(one == a[r]) r++; else break;
				}
				else if(l >= 1 && r > n) {
					one = min((T - now) / ((x[i] - x[l]) * 2), a[l]);
					now += (x[i] - x[l]) * 2 * one, ans += one;
					if(one == a[l]) l--; else break;
				}
			}
			maxn = max(maxn, ans + a[i]);
		}
		cout << maxn << '\n';
		return;
	}
}

int main() {
	n = read(), T = read();
	for(int i = 1; i <= n; i++) x[i] = read();
	for(int i = 1; i <= n; i++) a[i] = read();
	if(n <= 1000) return sub::Main(), 0;

	return 0;
}

#include<bits/stdc++.h>
#define int long long
using namespace std;

const int N = 5e5 + 11;

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
	return x * f;
}

int s[N], T;
int x[N], a[N], n;

bool check(int H) {
	int nowh = 0, tim = 0;
	int lc = 0, rc = 0, l = 1, r = n + 1;
	for (int i = 1; i <= n; i++)
		if (nowh + a[i] <= H) nowh += a[i], tim += (int)(x[i] - x[1]) * a[i];
		else {r = i, rc = H - nowh, tim += (int)(x[i] - x[1]) * rc; break;}
	if (tim <= T) return 1;
	for (int i = 2; i <= n; i++) {
		int suml = s[i - 1] - s[l - 1] - lc;
		int sumr = s[r - 1] - s[i - 1] + rc;
		tim += (x[i] - x[i - 1]) * (suml - sumr);
		while (r <= n && x[i] - x[l] > x[r] - x[i]) {
			int can = min(a[l] - lc, a[r] - rc);
			tim += (x[r] - x[i] - x[i] + x[l]) * can;
			lc += can, rc += can;
			if (lc >= a[l]) ++l, lc = 0;
			if (rc >= a[r]) ++r, rc = 0;
		}
		if (tim <= T) return 1;
	}
	return 0;
}

signed main() {
	n = read(), T = read(); T /= 2;
	for (int i = 1; i <= n; i++) x[i] = read();
	for (int i = 1; i <= n; i++) a[i] = read();
	for (int i = 1; i <= n; i++) s[i] = s[i - 1] + a[i];
	int l = 0, r = s[n], ans;
	while (l <= r) {
		int mid = (l + r) >> 1;
		if (check(mid)) l = mid + 1, ans = mid;
		else r = mid - 1;
	}
	cout << ans << '\n';
	return 0;
}

T3

神仙字符串题，不会

#include<bits/stdc++.h>
#define N 500010
#define ls (x << 1)
#define rs (x << 1 | 1)
using namespace std;
typedef long long ll;
struct intv{
	int s, i;
} a[N], b[N];
int p[N], rc[N << 2], f[N], n, m, type, anss, sum[N << 2];
ll ansf;
char s[N];
bool cmp(intv a, intv b) {
	return a.s < b.s;
}
inline void updata(int x, int l, int r, int p) {
	if (l == r) {
		rc[x] = r, sum[x] = 1; return;
	}
	int mid = (l + r) >> 1;
	if (p <= mid) updata(ls, l, mid, p);
	else updata(rs, mid + 1, r, p);
	sum[x] = sum[ls] + sum[rs];
	rc[x] = (rc[rs] == -1 ? rc[ls] : rc[rs]);
}
inline intv query(int x, int l, int r, int xl, int xr) {
	if (l == xl && r == xr) return (intv) {sum[x], rc[x]};
	int mid = (l + r) >> 1;
	if (xr <= mid) return query(ls, l, mid, xl, xr);
	else if (xl > mid) return query(rs, mid + 1, r, xl, xr);
	else {
		intv ql = query(ls, l, mid, xl, mid), qr = query(rs, mid + 1, r, mid + 1, xr);
		return (intv) {ql.s + qr.s, qr.i == -1 ? ql.i : qr.i};
	}
}
int main() {
	freopen("htstr.in", "r", stdin);
	freopen("htstr.out", "w", stdout);
	scanf("%s%d", s + 1, &type);
	n = strlen(s + 1); s[0] = '#';
	int mx = 0, id, now = 1;
	for (int i = 1; i <= n; i++) {
		if (mx >= i) p[i] = min(mx - i, p[2 * id - i]);
		for (; s[i + p[i] + 1] == s[i - p[i]]; p[i]++);
		if (p[i] + i < mx) id = i, mx = p[i] + i;
	}
	for (int i = 1; i <= n; i++) a[i] = (intv) {i - p[i], i};
	sort(a + 1, a + n + 1, cmp);
	for (int i = 1; i <= n << 2; i++) rc[i] = -1;
	for (int i = 1; i <= n; i++) {
		while (now <= n && a[now].s <= i) updata(1, 1, n, a[now].i), now++;
		if (p[i] <= 1) continue;
		intv qs = query(1, 1, n, i + 1, i + p[i] / 2);
		ansf += (ll)qs.s;
		if (qs.i != -1) b[++m] = (intv) {3 * i - 2 * qs.i + 1, 2 * qs.i - i};
	}
	if (type == 1) {printf("%I64d\n", ansf + n); return 0;}
	sort(b + 1, b + m + 1, cmp);
	id = 1, mx = 0;
	for (int r = 1; r <= n; r++) {
		while (b[id].s <= r && id <= m) mx = max(mx, b[id].i), id++;
		if (mx < r) ++anss, mx = r;
		else r = mx, ++anss;
	}
	if (type == 2) printf("%d\n", anss);
	else printf("%I64d %d\n", ansf + n, anss);
	return 0;
}