Codeforces Round #349

终于又回到熟悉的Round了

数学 A - Pouring Rain

设个未知数，解方程，还好没有hack点

#include <bits/stdc++.h>

typedef long long ll;
const int N = 1e5 + 5;
const double PI = acos (-1.0);

int main() {
    double d, h, v, e; scanf ("%lf%lf%lf%lf", &d, &h, &v, &e);
    double V = PI * (d / 2.0) * (d / 2.0) * h;
    double addv = PI * (d / 2.0) * (d / 2.0) * e;
    if (addv > v) {
        puts ("NO");
    } else {
        double t = -V / (addv - v);
        puts ("YES");
        printf ("%.8f
", t);
    }
    return 0;
}

数学 B - Coat of Anticubism

题意：求增加最小长度的一根木棍，使得构成一个多边形。

分析：那么构成三角形，原来n条木棍分成A，B两边，A和B接近(A<=B)，那么另一条边满足A + C > B，即C = B +１ - A

#include <bits/stdc++.h>

typedef long long ll;
const int N = 1e5 + 5;
int a[N];

int main() {
    int n; scanf ("%d", &n);
    ll sum = 0;
    for (int i=0; i<n; ++i) {
        scanf ("%d", a+i);
        sum += a[i];
    }
    std::sort (a, a+n);
    ll del = 1000001000;
    ll A, B, presum = 0;
    for (int i=0; i<n; ++i) {
        presum += a[i];
        ll res = sum - presum;
        ll tmp = presum - res;
        if (abs (tmp) < del) {
            if (tmp < 0) {
                del = -tmp;
                A = presum;
                B = res;
            } else {
                del = tmp;
                A = res;
                B = presum;
            }
        }
    }
    ll C = B + 1 - A;
    printf ("%I64d
", C);

    return 0;
}

DP C - Reberland Linguistics

题意：问后缀由若干个长度为2或3的子串构成，且相邻的子串不能相同。

分析：The only restriction — it is not allowed to append the same string twice in a row! 英语太渣不知道这是连续，相邻的意思。那么dp[i][0]表示i开始长度为2的子串能否可行，如果可行，那么dp[i-3][1]一定可行，因为长度不相等；还有如果长度相等的判断一下，即dp[i-2][0]。

#include <bits/stdc++.h>

const int N = 1e4 + 5;
bool dp[N][2];

int main() {
    std::string str;
    std::cin >> str;
    int n = str.length ();
    dp[n-2][0] = dp[n-3][1] = true;
    std::set<std::string> ans;
    for (int i=n-1; i>=5; --i) {
        if (dp[i][0]) {
            ans.insert (str.substr (i, 2));
            dp[i-3][1] = true;
            if (str.substr (i-2, 2) != str.substr (i, 2)) {
                dp[i-2][0] = true;
            }
        }
        if (dp[i][1]) {
            ans.insert (str.substr (i, 3));
            dp[i-2][0] = true;
            if (str.substr (i-3, 3) != str.substr (i, 3)) {
                dp[i-3][1] = true;
            }
        }
    }
    std::cout << ans.size () << '
';
    for (auto s: ans) {
        std::cout << s << '
';
    }
    return 0;
}

图论 D - World Tour

题意：找4个点按照顺序走，a->b->c->d，每次点到下一个点走的是最短路，问走的长度总和最大是多少。

分析：先计算dis(u, v)，枚举b和c，对于b来说在反向图中找距离最远的点，因为a!=b a!=c，所以存最优的前3个点；对于c来说在原图中找距离最远的点，存最优的前4个点。

#include <bits/stdc++.h>

const int N = 3e3 + 5;
const int M = 5e3 + 5;
const int INF = 0x3f3f3f3f;
int dis[N][N];
std::vector<int> G[N], rG[N];
std::vector<std::pair<int, int> > bin[N], bout[N];
bool vis[N];
int n, m;

void BFS() {
    memset (dis, INF, sizeof (dis));
    for (int i=1; i<=n; ++i) {
        std::queue<int> que;
        dis[i][i] = 0;
        que.push (i);
        while (!que.empty ()) {
            int u = que.front (); que.pop ();
            for (auto v: G[u]) {
                if (dis[i][v] > dis[i][u] + 1) {
                    dis[i][v] = dis[i][u] + 1;
                    que.push (v);
                }
            }
        }
    }
}

void sort_out() {
    for (int i=1; i<=n; ++i) {
        memset (vis, false, sizeof (vis));
        vis[i] = true;
        bout[i].push_back (std::make_pair (0, i));
        std::queue<std::pair<int, int> > que;
        que.push (std::make_pair (0, i));
        while (!que.empty ()) {
            std::pair<int, int> pu = que.front (); que.pop ();
            for (auto v: G[pu.second]) {
                if (!vis[v]) {
                    vis[v] = true;
                    bout[i].push_back (std::make_pair (pu.first + 1, v));
                    que.push (std::make_pair (pu.first + 1, v));
                }
            }
        }
        std::sort (bout[i].begin (), bout[i].end (), std::greater<std::pair<int, int> > ()); //dis[i][v], v
        if (bout[i].size () > 4) {
            bout[i].resize (4);
        }
    }
}

void sort_in() {
    for (int i=1; i<=n; ++i) {
        memset (vis, false, sizeof (vis));
        vis[i] = true;
        bin[i].push_back (std::make_pair (0, i));
        std::queue<std::pair<int, int> > que;
        que.push (std::make_pair (0, i));
        while (!que.empty ()) {
            std::pair<int, int> pu = que.front (); que.pop ();
            for (auto v: rG[pu.second]) {
                if (!vis[v]) {
                    vis[v] = true;
                    bin[i].push_back (std::make_pair (pu.first + 1, v));
                    que.push (std::make_pair (pu.first + 1, v));
                }
            }
        }
        std::sort (bin[i].begin (), bin[i].end (), std::greater<std::pair<int, int> > ()); //dis[v][i], v
        if (bin[i].size () > 3) {
            bin[i].resize (3);
        }
    }
}

int main() {
    scanf ("%d%d", &n, &m);
    for (int u, v, i=0; i<m; ++i) {
        scanf ("%d%d", &u, &v);
        G[u].push_back (v);
        rG[v].push_back (u);
    }
    BFS ();
    sort_in ();
    sort_out ();
    int a, b, c, d;
    int best = 0;
    for (int i=1; i<=n; ++i) {
        for (int j=1; j<=n; ++j) {
            if (i == j || dis[i][j] == INF) {
                continue;
            }
            int k, l;
            for (int ii=bin[i].size ()-1; ii>=0; --ii) {
                int tot = dis[i][j];
                std::pair<int, int> &x = bin[i][ii];
                if (x.second != i && x.second != j) {
                    k = x.second;
                    tot += x.first;
                    for (int jj=bout[j].size ()-1; jj>=0; --jj) {
                        std::pair<int, int> &y = bout[j][jj];
                        if (y.second != i && y.second != j && y.second != k) {
                            l = y.second;
                            tot += y.first;
                            if (best < tot) {
                                best = tot;
                                a = k; b = i; c = j; d = l;
                            }
                            tot -= y.first;
                        }
                    }
                }
            }
        }
    }
    printf ("%d %d %d %d
", a, b, c, d);

    return 0;
}

组合数学 C - Codeword(div 1)

题意：给一个串，长度为 l，问它扩展成长度n的串有多少个（其中长度l的原串相当于变成长度n的子序列）

分析：对于长度为 l 的子序列，n 的答案是

因为 l 的总和不超过 10^5​​，所以 l 不同的取值只有最多 2 sqrt(10^5)​​​​​ 种。对于每个 l，前缀和预处理所有 n 的答案。

令，则

#include <bits/stdc++.h>

typedef long long ll;
const int N = 1e5 + 5;
const int MOD = 1e9 + 7;
int fac[N], inv_fac[N];
char s[N];
int ans[N];
std::vector<std::pair<int, int> > query[N]; //len, n

int pow_mod(int x, int n) {
    int ret = 1;
    while (n) {
        if (n & 1) {
            ret = (ll) ret * x % MOD;
        }
        x = (ll) x * x % MOD;
        n >>= 1;
    }
    return ret;
}

int binom(int n, int m) {
    if (m > n) {
        return 0;
    } else {
        return (ll) fac[n] * inv_fac[m] % MOD * inv_fac[n-m] % MOD;
    }
}

void prepare(int len, int maxn) {
    int base = 1;
    //g(n)
    for (int i=0; i<=maxn; ++i) {
        ans[i] = i < len ? 0 : (ll) binom (i - 1, len - 1) * base % MOD;
        if (i >= len) {
            base = (ll) base * 25 % MOD;
        }
    }
    //f(n+1) = g(n+1) + f(n+1) * 26
    for (int i=0; i<maxn; ++i) {
        if (ans[i] >= MOD) {
            ans[i] -= MOD;
        }
        ans[i+1] += (ll) ans[i] * 26 % MOD;
    }
    if (ans[maxn] >= MOD) {
        ans[maxn] -= MOD;
    }
}

int main() {
    fac[0] = 1;
    for (int i=1; i<N; ++i) {
        fac[i] = (ll) fac[i-1] * i % MOD;
    }
    inv_fac[N-1] = pow_mod (fac[N-1], MOD-2);
    for (int i=N-2; i>=0; --i) {
        inv_fac[i] = (ll) inv_fac[i+1] * (i + 1) % MOD;
    }
    std::vector<int> query;
    int m; scanf ("%d", &m);
    scanf ("%s", s);
    int len = strlen (s);
    int maxn = -1;
    for (int i=0; i<m; ++i) {
        int op; scanf ("%d", &op);
        if (op == 1) {
            prepare (len, maxn);
            for (int j=0; j<query.size (); ++j) {
                printf ("%d
", ans[query[j]]);
            }
            query.clear ();
            maxn = -1;
            scanf ("%s", s);
            len = strlen (s);
        } else {
            int n; scanf ("%d", &n);
            if (maxn < n) {
                maxn = n;
            }
            query.push_back (n);
        }
    }
    prepare (len, maxn);
    for (int j=0; j<query.size (); ++j) {
        printf ("%d
", ans[query[j]]);
    }
    return 0;
}