目录

• Contest Info
• Solutions
  • A. Fifty-Fifty
  • B. Ordinary Number
  • C. Divide the Problems
  • D. Blue and Red Balls
  • E. Hopscotch Addict
  • F. Small Products

Contest Info

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Practice Link

Solved	A	B	C	D	E	F
6/6	O	O	O	O	Ø	O

• O 在比赛中通过
• Ø 赛后通过
• ! 尝试了但是失败了
• - 没有尝试

Solutions

---

A. Fifty-Fifty

#include <bits/stdc++.h>
using namespace std;

int main() {
    string s; cin >> s;
    sort(s.begin(), s.end());
    s.erase(unique(s.begin(), s.end()), s.end());
    cout << (((int)s.size() == 2) ? "Yes" : "No") << "\n";
    return 0;
}

B. Ordinary Number

#include <bits/stdc++.h>
using namespace std;

#define N 50
int n, a[N];

int main() {
    while (scanf("%d", &n) != EOF) {
        for (int i = 1; i <= n; ++i) {
            scanf("%d", a + i);
        }
        int res = 0;
        for (int i = 2; i < n; ++i) {
            int f = a[i] < a[i - 1];
            int g = a[i] < a[i + 1];
            if (f ^ g) {
                ++res;
            }
        }
        printf("%d\n", res);
    }
    return 0;
}

C. Divide the Problems

#include <bits/stdc++.h>
using namespace std;

#define N 100010
int n, a[N];

int main() {
    while (scanf("%d", &n) != EOF) {
        for (int i = 1; i <= n; ++i) {
            scanf("%d", a + i);
        }
        sort(a + 1, a + 1 + n);
        int mid = n / 2;
        if (a[mid] == a[mid + 1]) {
            puts("0");
        } else {
            printf("%d\n", a[mid + 1] - a[mid]);
        }
    }
    return 0;
}

D. Blue and Red Balls

题意：
有\(K\)个蓝球，\(N - K\)个红球，询问将蓝球分成\(i(1 \leq i \leq k)\)堆，中间用红球隔开的方案数分别是多少?

思路：
考虑先将\(k\)个蓝球分成\(i\)堆，然后每堆后面跟着一个红球，然后就有\(i + 1\)个空隙，剩下的红球相当于要放入\(i + 1\)个空箱中，允许空箱的经典问题。

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 2010
const int p = 1e9 + 7;
int n, k;
ll fac[N], inv[N];
ll qmod(ll base, ll n) {
    ll res = 1;
    while (n) {
        if (n & 1) {
            res = res * base % p;
        }
        base = base * base % p;
        n >>= 1;
    }
    return res;
}

ll C(int n, int m) {
    if (n < m) return 0;
    return fac[n] * inv[m] % p * inv[n - m] % p;
}

ll f(int n, int i, int k) {
    ll remind = n - k;
    if (n - k < i - 1) {
        return 0;
    }
    remind -= i - 1;
    return C(k - 1, i - 1) * C(i + remind, i) % p; 
}

int main() {
    fac[0] = 1;
    for (int i = 1; i <= 2000; ++i) {
        fac[i] = fac[i - 1] * i % p;
    }
    inv[2000] = qmod(fac[2000], p - 2);
    for (int i = 2000; i >= 0; --i) {
        inv[i - 1] = inv[i] * i % p;
    }
    while (scanf("%d%d", &n, &k) != EOF) {
        for (int i = 1; i <= k; ++i) {
            printf("%lld\n", f(n, i, k));
        }
    }
    return 0;
}

E. Hopscotch Addict

题意：
求\(S\)到\(T\)的最短路，并且要满足其长度是\(3\)的倍数。

思路：
\(dis[i][j]\)表示到第\(i\)个点，长度模\(3\)的值为\(j\)的最短路是多少，然后跑Dijkstra即可。

#include <bits/stdc++.h>
using namespace std;

#define INF 0x3f3f3f3f
#define N 100010
int n, m, s, t;
vector <vector<int>> G;
struct node {
    int u, w;
    node() {}
    node (int u, int w) : u(u), w(w) {}
    bool operator < (const node &other) const {
        return w > other.w;
    }
};

int dis[N][3], used[N][3];
void BFS() {
    for (int i = 1; i <= n; ++i) {
        for (int j = 0; j < 3; ++j) {
            dis[i][j] = INF;
            used[i][j] = 0;     
        }
    }
    dis[s][0] = 0;  
    priority_queue <node> pq;
    pq.push(node(s, 0)); 
    while (!pq.empty()) {
        int u = pq.top().u, w = pq.top().w; pq.pop(); 
        if (used[u][w % 3]) continue;
        used[u][w % 3] = 1;
        for (auto v : G[u]) {
            if (dis[v][(w + 1) % 3] > w + 1) {
                dis[v][(w + 1) % 3] = w + 1;  
                pq.push(node(v, w + 1));  
            }
        }
    }
}

int main() {
    while (scanf("%d%d", &n, &m) != EOF) {
        G.clear(); G.resize(n + 1);
        for (int i = 1, u, v; i <= m; ++i) {
            scanf("%d%d", &u, &v);
            G[u].push_back(v);
        }
        scanf("%d%d", &s, &t);
        BFS();
        if (dis[t][0] == INF) puts("-1");
        else {
            printf("%d\n", dis[t][0] / 3);
        }
    }
    return 0;
}

F. Small Products

题意：
构造一个长度为\(k\)的，并且相邻两数之积不超过\(N\)的序列的方案数是多少？

思路：
\(f[i][j]\)表示到第\(i\)位，当前位为\(j\)的方案数是多少。
暴力\(DP\)显然不行。
但是我们考虑我们每次对于\(j\)转移的都是\(\sum\limits_{j = 1}^{N / j} f[i - 1][j]\)，考虑到\(N / j\)的取值只有\(2\sqrt{N}\)种，所以只需要存\(k \cdot 2\sqrt{N}\)种状态。转移即可。

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 110
#define M 100010
#define pii pair <int, int>
#define fi first
#define se second
const ll p = 1e9 + 7;
int n, k;
ll f[N][M];
pii g[M]; 
map <int, int> mp;

int main() {
    while (scanf("%d%d", &k, &n) != EOF) {
        memset(f, 0, sizeof f);
        mp.clear();
        int m = 0;
        for (int i = 1, j; i <= k; i = j + 1) {
            j = k / (k / i);
            g[++m] = pii(j, j - i + 1);
            mp[j] = m;
        }
        for (int i = 1; i <= m; ++i) {
            f[1][i] = (f[1][i - 1] + g[i].se) % p; 
        }
        for (int i = 2; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                f[i][j] = (f[i][j - 1] + 1ll * g[j].se * f[i - 1][mp[k / g[j].fi]] % p) % p;
            }
        }
        printf("%lld\n", f[n][m]);
    }
    return 0;
}