2020蓝桥杯国赛A组c++

A

签到

int main() {
    for (int i = 2; i <= 2020; ++i) {
        if (!v[i]) prime[++tot] = i;
        for (int j = 1; j <= tot && prime[j] <= 2020 / i; ++j) {
            v[i * prime[j]] = 1;
            if (i % prime[j] == 0) break;
        }
    }
    cout << 2020 - 1 - tot;
    return 0;
}

B

我sb, 没算年份是2的, 还搁那里沾沾自喜, 算了闰年的2月, 臭sb

using ll = long long;

bool check(int x) {
    while (x) {
        if (x % 10 == 2) return 1;
        x /= 10;
    }
    return 0;
}

int main () {
    ll ans = 0;
    rep (i, 1900, 9999) {
        if (i % 100 == 0 && i % 400 == 0) ans += 1;//闰年多一天
        else if (i % 100 && i % 4 == 0) ans += 1;//闰年多一天
        if (check(i)) ans += 365;//年份带2,直接+365
        else ans += 179;//除了(2,12)其他每个月只有(2,12,20~29), 即10*12 + 28 + 31
    }
    cout << ans;
    return 0;
}

C

O(n^2)

int main() {
    string s; cin >> s;
    set<string> st;
    for (int i = 0; i < s.size(); ++i) {
        string t = "";
        for (int j = i; j < s.size(); ++j) {
            if (t == "" || s[j] > t.back()) t += s[j];
            else break;
            st.insert(t);
        }
    }
    cout << st.size();
    //tocyjkdzcieoiodfpbgcncsrjbhmugdnojjddhllnofawllbhfiadgdc
    //djstemphmnjihecoapdjjrprrqnhgccevdarufmliqijgihhfgdcmxvi
    //cfauachlifhafpdccfseflcdgjncadfclvfmadvrnaaahahndsikzssoy
    //wakgnfjjaihtniptwoulxbaeqkqhfwl
    return 0;
}

D

看大题的代码, 自己慢慢暴力跑吧

E

爆搜

struct node {
    int s[16];
    bool operator <(const node b) const {
        for (int i = 0; i < 16; ++i)
            if (s[i] != b.s[i]) return s[i] < b.s[i];
        return false;
    }
} c;

bool v[16];
int d[4][2] = { {1, 0}, {-1, 0}, {0, 1}, {0, -1} };
set<node> st;

bool check(int x, int y) {
    if (x < 0 || x > 3 || y < 0 || y > 3) return 1;
    return v[x * 4 + y];
}

void dfs(int x, int y, int k, node& c) {
    if (k == 16) { st.insert(c); return; }
    for (int i = 0; i < 4; ++i) {
        int nx = x + d[i][0], ny = y + d[i][1];
        if (check(nx, ny)) continue;
        c.s[k] = nx * 4 + ny; v[nx * 4 + ny] = 1;
        dfs(nx, ny, k + 1, c);
        v[nx * 4 + ny] = 0;
    }
}

int main() {
    for (int i = 0; i < 16; ++i) {
        v[i] = 1; c.s[0] = i;
        dfs(i / 4, i % 4, 1, c);
        v[i] = 0;
    }
    cout << st.size();
    return 0;
}

F

我用long double 精度会丢失, 还是不是正解

要么写高精

要么看题目, 保证答案在1e9内, 那就要在递归的时候考虑更多的东西, 不能直接记步数

而是记录你走了 3 的几次方 的步, 有点像倍增, 但是记录的基地为3 而不是 2,

最后根据两个坐标点走过的 (3^i) 的数量做差, 在记录差和, ((3^i) = (3 * 3^{i-1})), 不断自顶向下做差, 最后答案在1e9内

代码就不写了, 就在原基础上改改就行

---

以下为原思路

跟分型之城有点像, 不过这题分9块

而且是 (3^{100} * 3^{100}), 要开long double

int tt[3][3] = {
    {1, 6, 7},
    {2, 5, 8},
    {3, 4, 9}
};

long double fac[205];

long double solve(long long x, long long y, long long k) {
    if (k == 1) return tt[x][y];
    int idx = x / fac[k - 1], idy = y / fac[k - 1];
    int id = idx * 3 + idy;
    switch (id) {
        case 0: return solve(x, y ,k - 1);
        case 1: return fac[k - 1 << 1] * 5 + solve(fac[k - 1] - 1 - x, y - fac[k - 1], k - 1);
        case 2: return fac[k - 1 << 1] * 6 + solve(x, y - 2 * fac[k - 1], k - 1);
        case 3: return fac[k - 1 << 1] + solve(x - fac[k - 1], fac[k - 1] - 1 - y, k - 1);
        case 4: return fac[k - 1 << 1] * 4 + solve(2 * fac[k - 1] - 1 - x, 2 * fac[k - 1] - 1 - y, k - 1);
        case 5: return fac[k - 1 << 1] * 7 + solve(x - fac[k - 1], fac[k] - 1 - y, k - 1);
        case 6: return fac[k - 1 << 1] * 2 + solve(x - 2 * fac[k - 1], y, k - 1);
        case 7: return fac[k - 1 << 1] * 3 + solve(fac[k] - 1 - x, y - fac[k - 1], k - 1);
        case 8: return fac[k - 1 << 1] * 8 + solve(x - 2 * fac[k - 1], y - 2 * fac[k - 1], k - 1);
        default:break;
    }
}

int main() {
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    long long k, x, y, a, b; cin >> k >> y >> x >> b >> a;
    fac[0] = 1;
    for (int i = 1, j = k << 1; i <= j; ++i) fac[i] = fac[i - 1] * 3;
    long long ans = fabs(solve(x, y, k) - solve(a, b, k));
    cout << ans;
    return 0;
}

G

搁着阅读理解呢? 太长不看, 输出样例结束

H

国王硬币类似, 证明就不给了(懒

pair<ll, ll> s[1001];


int main() {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        ll a, b, c; cin >> a >> b >> c;
        s[i].first = a + b + c; s[i].second = a + b;
    }
    sort(s + 1, s + 1 + n);
    ll ans = 0, cur = 0;
    for (int i = 1; i <= n; ++i) {
        ans += cur + s[i].second;
        cur += s[i].first;
    }
    cout << ans;
    return 0;
}

I

扫描线的板子题稍微改改

const int N = 2e5 + 5;

struct BIT {
    struct node {
        int l, r;
        ll cnt, a, b, len;
    } t[N << 2];

    void build(int rt, int l, int r, vector<int>& st) {
        t[rt].l = l, t[rt].r = r, t[rt].len = st[r] - st[l];
        if (l + 1 == r) return;
        int mid = l + r >> 1;
        build(rt << 1, l, mid, st);
        build(rt << 1 | 1, mid, r, st);
    }

    void change(int rt, int l, int r, int k) {
        if (t[rt].l >= l && t[rt].r <= r && t[rt].l + 1 == t[rt].r) {
            t[rt].cnt += k;
            if (t[rt].cnt & 1) t[rt].a = t[rt].len, t[rt].b = 0;
            else if (t[rt].cnt) t[rt].b = t[rt].len, t[rt].a = 0;
            else t[rt].a = t[rt].b = 0;
            return;
        }
        int mid = t[rt].l + t[rt].r >> 1;
        if (mid > l) change(rt << 1, l, r, k);
        if (mid < r) change(rt << 1 | 1, l, r, k);
        t[rt].a = t[rt << 1].a + t[rt << 1 | 1].a;
        t[rt].b = t[rt << 1].b + t[rt << 1 | 1].b;
    }

    ll ask(bool k) {
        return k ? t[1].a : t[1].b; 
    }
} T;

int n;
vector<int> st;
pair<ll, pair<pair<ll, ll>, int> > lin[N];

int get(int x) {
    return lower_bound(st.begin(), st.end(), x) - st.begin();
}

bool cmp(pair<ll, pair<pair<ll, ll>, int> > a, pair<ll, pair<pair<ll, ll>, int> > b) {
    if (a.fi != b.fi) return a.fi < b.fi;
    return a.se.se > b.se.se;
}

int main() {
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    cin >> n;
    for (int i = 0; i < n; ++i) {
        int x, y, a, b; cin >> x >> y >> a >> b;
        if (x > a) swap(a, x);
        if (y > b) swap(y, b);
        lin[i << 1] = { y, { { x, a }, 1 } };
        lin[i << 1 | 1] = { b, { { x, a}, -1 } };
        st.push_back(x); st.push_back(a);
    }
    st.push_back(-2e9);
    sort(st.begin(), st.end());
    st.erase(unique(st.begin(), st.end()), st.end());
    sort(lin, lin + (n << 1), cmp);
    T.build(1, 1, st.size() - 1, st);
    T.change(1, get(lin[0].se.fi.fi), get(lin[0].se.fi.se), lin[0].se.se);
    ll a = 0, b = 0;
    for (int i = 1; i < n << 1; ++i) {
        ll h = lin[i].fi - lin[i - 1].fi;
        a += T.ask(1) * h; b += T.ask(0) * h;
        T.change(1, get(lin[i].se.fi.fi), get(lin[i].se.fi.se), lin[i].se.se);
    }
    cout << a << '
' << b;
    return 0;
}

J

数据太大了, 直接写了个假的 dp 骗分, 就不写了, 又不是正解