2018CCPC吉林赛区

A

分块, 签到

int main() {
    IOS; int cas = 0;
    for (cin >> _; _; --_) {
        cin >> n; __int128 cur = 0;
        for (int i = 1, j; i <= n; i = j + 1) {
            j = min(n, n / (n / i));
            cur += (__int128)n / i * (j - i + 1);
        }
        cout << "Case " << ++cas << ": ";
        if (cur & 1) cout << "odd
";
        else cout << "even
";
    }
    return 0;
}

B

模拟

int main() {
    int cas = 0;
    for (scanf("%d", &_); _; --_) {
        int a, b; scanf("%d:%d %s", &a, &b, s + 1);
        if (s[1] == 'A' && a == 12) a = 0;
        else if (s[1] == 'P' && a != 12)  a += 12;

        scanf("%s%s", x + 1, y + 1);
        if (x[1] == 'B') a = a - 8;
        else if (x[1] == 'W') a = a + 5;
        else if (x[1] == 'M') a = a - 3;
        if (y[1] == 'B') a += 8;
        else if (y[1] == 'W') a -= 5;
        else if (y[1] == 'M') a += 3;

        cout << "Case " << ++cas << ": ";
        if (a >= 24) cout << "Tomorrow ";
        else if (a < 0) cout << "Yesterday ";
        else cout << "Today ";
        a = (a + 24) % 24;

        char c, d;
        if (b < 10) c = '0';
        else c = '0' + b / 10;
        d = '0' + b % 10; 
        if (a == 0 || a == 24) cout << "12:" << c << d << " AM
";
        else if (a < 12) cout << a << ":" << c << d << " AM
";
        else if (a == 12) cout << "12:" << c << d << " PM
";
        else cout << a - 12 << ':' << c << d << " PM
";
    }
    return 0;
}

C

不断合并就完事

int n, m, _, k;
int f[N], a[N];

int find(int x) {
    return x == f[x] ? x : f[x] = find(f[x]);
}

void unit(int x, int y) {
    x = find(x), y = find(y);
    if (x == y) return;
    f[y] = x;
}

int main() {
    IOS; int cas = 0;
    for (cin >> _; _; --_) {
        cin >> n; unordered_map<int, int> st;
        int x = 0, y = 0;
        rep (i, 1, n) {
            cin >> m, f[i] = i;
            while (st[m]) unit(i, st[m]), st[m--] = 0;
            if (m == 1) {
                if (x == 0) x = i;
                else if (y == 0) y = i;
            }
            else st[m] = i;
        }
        cout << "Case " << ++cas << ": ";
        if (x == 0 || y == 0) cout << "NO
";
        else {
            cout << "YES
";
            rep (i, 1, n) cout << (find(i) == x);
            cout << '
';
        }
    }
    return 0;
}

D

正解没写出来, 不过我dp倒是写出来了, 不过跑 1~100超时, 大概5~7s

但是是 1~100 呦~, 打表啊~

打表的dp(输入100就行)

int n, m, _, k;
ll c[4005][4005];
long double p, q, f[4005] = {1};

void init(int n) {
    rep (i, 0, n) c[i][0] = 1;
    rep (i, 1, n) rep (j, 1, i) c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
}

int main() {
    IOS; int cas = 0; init(4000);
    for (cin >> _; _; --_) {
         p = (100 - _ + 1) / 100.0;
         long double res = p * 0.02;
         rep (i, 0, 4000) f[i] = 0; f[1] = 1;
         rep (i, 1, 4000) {
             per (j, i, 0) {
                if (j) f[j] = f[j] * (1 - p) + f[j - 1] * p * (1 - min(j * 0.005 + 0.015 * i, 1.0));
                else f[j] = f[j] * (1 - p); 
                res += f[j] * p * min(1.0, 0.005 * j + 0.015 * i + 0.02) * (i + 1);
             }
         }
         cout << setiosflags(ios::fixed) << setprecision(12) << res << ',' << '
';
    }
    return 0;
}

表就自己打吧, 不贴了, 主要是暴力一直跑不出来, 没办法写了dp, 结果还不是正解, 鬼知道是公式啊, 反正打表也相当快(5~7s),单询问绝不会超时

E

设碰撞点为 ((x_1, y_1, z_1)), 则

(left{egin{matrix}x_1=x_0+v_x*t\y_1=y_0+v_y*t\z_1=z_0+v_z*tend{matrix} ight.)

考虑个相似三角形

$ frac{sqrt{x^2+y2}}{r} = frac {h-z_1}{h} $

拆开方, 带入原数, 得

((v_x^2+v_y^2-frac{r^2}{h^2}v_z^2)t^2+2(x*v_x+y*v_y-(z-h)frac{r^2}{h^2}v_z)t=frac{r^2}{h^2}*(h-z)^2-x^2-y^2)

即 (at^2+2bt=c)

解为 (t=frac{pm sqrt {b^2+ac}-b}{a})

然后就是判两个解那个是实根且能在最短时间碰到圆锥

double r, h, x, y, z, vx, vy, vz;

bool check(double t) {
    double a = x + vx * t, b = y + vy * t, c = z + vz * t;
    return fabs(sqrt(sqr(a) + sqr(b)) * h - (h - c) * r) < 1e-8;
}

int main() {
    IOS; int cas = 0;
    for (cin >> _; _; --_) {
        cin >> r >> h >> x >> y >> z >> vx >> vy >> vz;
        double a = sqr(vx) + sqr(vy) - sqr(r / h * vz);
        double b = x * vx + y * vy - sqr(r / h) * (z - h) * vz;
        double c = sqr(r / h * (h - z)) - sqr(x) - sqr(y);
        double der = sqrt(sqr(b) + a * c), t1 = (der - b) / a, t2 = (-der - b) / a;
        double t = 2e18;
        if (check(t1)) umin(t, t1);
        if (check(t2)) umin(t, t2); 
        cout << "Case " << ++cas << ": " << setiosflags(ios::fixed) << setprecision(12) << t << '
';
    }
    return 0;
}

F

签到题

int main() {
    IOS; int cas = 0;
    for (cin >> _; _; --_) {
        cin >> n; int ans = 0;
        rep (i, 1, n) {
            int r; cin >> r;
            if (r > 2) ans ^= r - 2;
        }
        cout << "Case " << ++cas << ": " << ans << '
';
    }
    return 0;
}

I

大模拟

PII a[N], b[N];

ll solve(bool f) {
    ll ans = 0; int i, j;
    for (i = 1, j = n; j && i <= m; ++i, --j) {
        if (a[j].se) { --i; continue; }
        if (b[i].se) { ++j; continue; }
        if (b[i].fi >= a[j].fi) return ans;
        ans += a[j].fi - b[i].fi;
    }
    if (f) while (j) if (!a[j--].se) ans += a[j + 1].fi;
    return ans;
}

int main() {
    IOS; int cas = 0;
    for (cin >> _; _; --_) {
        cin >> n >> m;
        rep (i, 1, n) cin >> a[i].fi, a[i].se = 0;
        rep (i, 1, m) cin >> b[i].fi;
        rep (i, 1, m) cin >> b[i].se;
        sort(a + 1, a + 1 + n); sort(b + 1, b + 1 + m);
        ll ans = solve(0);

        bool f = 1;
        for (int i = 1; i <= m && f; ++i)
            if (b[i].se) {
                int c = lower_bound(a + 1, a + 1 + n, PII{ b[i].fi, 0 }) - a;
                while (c <= n && a[c].se) ++c;
                if (c > n) f = 0;
                a[c].se = 1;
            }

        if (f) ans = max(ans, solve(1));
        cout << "Case " << ++cas << ": " << ans << '
';
    }
    return 0;
}