省选测试34

A 老夫

题目大意 ： n个人，如果能忍受广告就广告，如果能支付的起就交钱，问商家在每个播放广告数是能获得的最大利益

• 按忍受广告数量从小到大排序，那么就是前面的人去交钱或是推出，后面的人看广告

• 不看广告的人如果从大到小排序后，交钱的就是一段前缀，这样，贡献就是 (a_ii) 的最大值

• 以a为下标，分块维护凸包，找最大值的时候单调队列即可

Code

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#include <cmath>
#include <cstdio>
#include <algorithm>

using namespace std;
typedef long long ll;
const int N = 1e5 + 5, M = 320;

int read(int x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar()) if (c == '-') f = -1;
    for (; c >='0' && c <='9'; c = getchar()) x = x * 10 + c - '0';
    return x * f;
}

int n, k, ma, mb, a[N], b[N], p[N];
int c[N], tag[M], R[M], q[M][M], l[M], r[M];
ll w[N], ans;

bool Cmp(int x, int y) {
    return b[x] < b[y];
}

ll Cal (int x) {
    return w[x] + 1ll * tag[c[x]] * x;
}

bool Judge(int x, int y, int z) {
    return (w[y] - w[x]) * (z - x) < (w[z] - w[x]) * (y - x);
}

void Solve(int x) {
    int cx = c[x];
    for (int i = 1; i < cx; ++i) {
        tag[i]++;
        while (l[i] < r[i] && Cal(q[i][l[i]]) < Cal(q[i][l[i]+1])) l[i]++;
        ans = max(ans, Cal(q[i][l[i]]));
    }
    int &tp = r[cx] = 0; l[cx] = 1;
    for (int i = R[cx-1] + 1; i <= R[cx]; ++i) {
        w[i] += 1ll * tag[cx] * i;
        if (i <= x) w[i] += i;
        ans = max(ans, w[i]);
        while (tp && w[i] >= w[q[cx][tp]]) tp--;
        while (tp > 1 && Judge(q[cx][tp-1], q[cx][tp], i)) tp--;
        q[cx][++tp] = i;
    }
    tag[cx] = 0;
}

int main() {
    freopen("laofu.in", "r", stdin);
    freopen("laofu.out", "w", stdout);
    n = read(), k = read();
    for (int i = 1; i <= n; ++i) {
        a[i] = read(); b[i] = read(); p[i] = i;
        ma = max(ma, a[i]), mb = max(mb, b[i]); 
    }
    int sq = sqrt(ma);
    for (int i = 1; i <= ma; ++i) {
        c[i] = (i - 1) / sq + 1;
        R[c[i]] = i; l[c[i]] = 1;
        q[c[i]][++r[c[i]]] = i;
    }
    sort(p + 1, p + n + 1, Cmp);
    for (int i = 1, j = 1; i <= mb + 1; ++i) {
        for (; j <= n && b[p[j]] < i; ++j) Solve(a[p[j]]);
        printf("%lld ", ans + 1ll * k * i * (n - j + 1));
    }
    return 0;
}

B 打算

题目大意 ： 机器人会按照一段指令循环的运动，给出n个时刻观测到的位置，问这段指令可能是什么

• 把坐标系旋转45度，让(x,y)变为(x+y,x-y)这样每次移动横纵坐标都会改变，就可以分开计算

• 然后推个式子，找到一个合法的解就行了

Code

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#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#define double long double

using namespace std;
typedef long long ll;
const int N = 2e6 + 5;

ll read(ll x = 0, int f = 1, char c = getchar()) {
    for (; c < '0' || c > '9'; c = getchar()) if (c == '-') f = -1;
    for (; c >='0' && c <='9'; c = getchar()) x = x * 10 + c - '0';
    return x * f;
}

int n, m;
ll t[N], x[N], y[N];

struct Node {
    ll t, a, b;
}a[N];

bool operator < (const Node &x, const Node &y) {
    return x.t < y.t;
}

void Solve(ll *x) {
    ll l = -1e18, r = 1e18;
    for (int i = 1; i <= n; ++i)
        a[i] = (Node) {t[i] % m, -t[i] / m, x[i]};
    sort(a + 1, a + n + 1); a[n+1] = (Node) {m, 1, 0};
    for (int i = 0; i <= n; ++i) {
        ll T = a[i+1].t - a[i].t, A = a[i+1].a - a[i].a, B = a[i+1].b - a[i].b;
        if (!A) {
            if (abs(B) > T) { puts("NO"), exit(0); } 
            else continue;
        }
        double L = (double)(-T - B) / A, R = (double)(T - B) / A;
        if (L > R) swap(L, R);
        l = max(l, (ll)ceil(L)); r = min(r, (ll)floor(R));
    }
    if (l+m & 1) l++;
    if (r+m & 1) r--;
    if (l > r) { puts("NO"), exit(0); }
    int cnt = 0;
    for (int i = 0; i <= n; ++i) {
        a[i+1].b += a[i+1].a * l;
        int tmp = a[i+1].b - a[i].b;
        while (tmp > 0) x[++cnt] = 1, tmp--;
        while (tmp < 0) x[++cnt] = 0, tmp++;
        if (abs(a[i+1].t - cnt) & 1) { puts("NO"), exit(0); }
        while (cnt < a[i+1].t) x[++cnt] = 1, x[++cnt] = 0;
    }
}

int main() {
    freopen("dasuan.in", "r", stdin);
    freopen("dasuan.out", "w", stdout);
    n = read(); m = read();
    for (int i = 1; i <= n; ++i) {
        t[i] = read();
        ll a = read(), b = read();
        x[i] = a + b; y[i] = a - b;
    }
    Solve(x); Solve(y);
    for (int i = 1; i <= m; ++i)
        putchar(x[i] ? (y[i] ? 'R' : 'U') : (y[i] ? 'D' : 'L'));
    return 0;
}

C (Unaccepted)

题目大意 ：

Code

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