QBXT模擬賽2

總結

指望得分：\(100 + 40 + 0 = 140\)
實際得分：\(0 + 0 + 0 = 0\)

鬼知道爲何我代碼沒有交上。。自測\(10 + 50 + 0\)……這是心態爆炸的一場考試
\(T1\)敲了\(3\)個小時，寫了\(9kb\)，連樣例都過不了，改了改過了樣例迅速離開
\(T2\)只會\(40\)，但由於一個點的\(a[i]\)很大，因而水過去了，變成了\(50\)
\(T3\)連看都沒看，\(0\)分滾粗

我果是個彩筆

思路&&代碼

T1

模擬題， 一看這長長的題面，就知道是個簡單題（fp嘛我作了三個小時不過樣例能是簡單題？）

其實我以爲最難的不是別的，就是輸入。。。這麼一大堆字符串，還有\(5\)個位置，仍是兩個隊，如何作到最簡？

其實這些能夠化做一個總體，咱們很容易想到要用結構體存儲，但複雜的hapigou就是：結構體數組怎麼開？

咱們能夠開一個三維的\(a\)數組，存儲兩個隊全部位置全部人的信息，\(a[0]\)就表示\(A\)隊，\(a[1]\)就表示\(B\)隊，\(a[x][1-5]\)就表示\(x\)隊五個位置的信息，\(a[x][y][cnt]\)就表示\(x\)隊第\(y\)個位置的第\(cnt\)名選手，這樣不只好理解，還方便了以後咱們要用到的的排序

再者，確定要從實力由高到低的選，每次某個位置換的確定是這個位置剩餘的實力最強的人，因此咱們就能夠直接按照實力把每一個位置的人排序，若是實力相同則編號較小的在前

以後咱們就要想怎麼來選人了，要想讓場上全部人的綜合實力最大，咱們要保證換下來的人與在場的人實力的差值最小，因此就能夠在五個位置中找差值最小的位置來換，這樣這道題就完美解決了

#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

這道題到如今我仍是隻會五十分，很容易想到是一個貪心的思路，咱們直接用兩個指針表示從左邊右邊拿，就能拿五十了

滿分的也在下面，仍是不會。。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

神仙字符串題，不會

#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;   
}