2020 Multi-University Training Contest 6

Road To The 3rd Building

期望 = 区间的权值 * 区间出现的次数

当 $n=3$

$$E(x)=\sum x\times p(x)=\frac{a_1+\frac{1}{2}(a_1+a_2)+\frac{1}{3}(a_1+a_2+a_3)+a_2+\frac{1}{2}(a_2+a_3)+a_3}{\frac{n(n+1)}{2}}$$

$$=\frac{a_1(1+\frac{1}{2}+\frac{1}{3})+a_2(1+2\frac{1}{2}+\frac{1}{3})+a_3(1+\frac{1}{2}+\frac{1}{3})}{6}$$

当 $n=4$

$$E(x)=\frac{a_1+\frac{1}{2}(a_1+a_2)+\frac{1}{3}(a_1+a_2+a_3)+\frac{1}{4}(a_1+a_2+a_3+a_4)+a_2+\frac{1}{2}(a_2+a_3)+\frac{1}{3}(a_2+a_3+a_4)+a_3+\frac{1}{2}(a_3+a_4)+a_4}{10}$$

$$=\frac{a_1(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+a_2(1+2\frac{1}{2}+2\frac{1}{3}+\frac{1}{4})+a_3(1+2\frac{1}{2}+2\frac{1}{3}+\frac{1}{4})+a_4(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})}{10}$$

以此类推，当$n=5$

$$E(x)=\frac{a_1(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5})+a_2(1+2\frac{1}{2}+2\frac{1}{3}+2\frac{1}{4}+\frac{1}{5})+a_3(1+2\frac{1}{2}+3\frac{1}{3}+2\frac{1}{4}+\frac{1}{5})+a_4(1+2\frac{1}{2}+2\frac{1}{3}+2\frac{1}{4}+\frac{1}{5})+a_5(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5})}{15}$$

#include 
using namespace std;

typedef long long ll;
const ll mod = 1e9 + 7;
const int N = 1e6 + 10;
ll n;

ll qpow(ll a, ll b) {
    a %= mod;
    ll res = 1;
    while (b) {
        if (b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}


ll Add(ll x, ll y) {
    return (x + y) % mod;
}

ll Sub(ll x, ll y) {
    return (x - y + mod) % mod;
}

ll Mul(ll x, ll y) {
    return (x * y) % mod;
}

ll demo[N], sum[N], a[N];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    for (int i = 1; i <= 200010; i++) {
        demo[i] = qpow(i, mod - 2); // 1/i
        sum[i] = Add(sum[i - 1], demo[i]); // ∑1/i
    }

    int T;
    cin >> T;
    while (T--) {
        cin >> n;
        for (int i = 1; i <= n; i++) {
            cin >> a[i];
        }

        ll tmp = 0;
        ll res = 0;
        for (int i = 1, j = n; i <= j; i++, j--) {
            // tmp+= sum[j]-sum[i-1]
            tmp = Add(tmp, Sub(sum[j], sum[i - 1]));

            // res+=a[i] * tmp + a[n - i + 1] * tmp;
            res = Add(res, Mul(a[i], tmp));
            if (j > i) {
                res = Add(res, Mul(a[j], tmp));
            }
        }

        ll Inv = qpow(n * (n + 1) / 2, mod - 2); // 1/区间个数
        res = Mul(res, Inv);
        cout << res << endl;
    }
    return 0;
}

---

Little Rabbit’s Equation

模拟

#include 
using namespace std;
#define between(x, a, b) (a<=x && x<=b)
typedef long long ll;
int bit[20];
string num1, num2, num3;

ll toInt(string s, int b) {
    int sz = s.length();
    for (int i = 0; i < sz; i++) {
        bit[i] = (between(s[i], '0', '9') ? s[i] - '0' : s[i] - 'A' + 10);
    }

    ll res = 0;
    for (int i = 0; i < sz; i++) {
        if (bit[i] >= b) return -1; // 判断数字是否合法
        res = res * b + bit[i];
    }
    return res;
}

bool solve(ll a, ll b, ll c, int op) {
    if (op == 1) return (a + b) == c;
    else if (op == 2) return (a - b) == c;
    else if (op == 3) return (a * b) == c;
    else return ((a % b) == 0 && (a / b == c)); // 防整除精度问题
}

bool check(int b, int op) {
    ll x = toInt(num1, b);
    if (x == -1) return false;
    
    ll y = toInt(num2, b);
    if (y == -1) return false;
    
    ll z = toInt(num3, b);
    if (z == -1) return false;
    
    return solve(x, y, z, op);
}

string s;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    while (cin >> s) {
        int sz = s.length();
        int i = 0, j, op;

        while (i < sz && (between(s[i], '0', '9') || between(s[i], 'A', 'F'))) i++;
        num1 = s.substr(0, i);

        if (s[i] == '+') op = 1;
        else if (s[i] == '-') op = 2;
        else if (s[i] == '*') op = 3;
        else if (s[i] == '/') op = 4;

        i++;
        j = i;
        while (i < sz && (between(s[i], '0', '9') || between(s[i], 'A', 'F'))) i++;
        num2 = s.substr(j, i - j);

        i++;
        j = i;
        while (i < sz && (between(s[i], '0', '9') || between(s[i], 'A', 'F'))) i++;
        num3 = s.substr(j, i - j);

        int f = -1;
        for (int i = 2; i <= 16; i++) {
            if (check(i, op)) {
                f = i;
                break;
            }
        }
        cout << f << endl;
    }
    return 0;
}

Borrow
Asteroid in Love
Fragrant numbers

A Very Easy Graph Problem

官方题解

树上任意两点间距离的改版

#include 
using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
const int N = 1e6 + 10;

struct Edge {
    int v;
    ll w;
};
vector<Edge> e[N];

int a[N];
int n, m;
int fa[N];

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

void Union(int x, int y, ll w) {
    int fx = find(x);
    int fy = find(y);
    if (fx == fy) return;

    e[x].push_back({y, w});
    e[y].push_back({x, w});

    if (fx < fy) {
        fa[fy] = fx;
    } else {
        fa[fx] = fy;
    }
}

ll Mul(ll x, ll y) {
    return (x * y) % mod;
}

ll Add(ll x, ll y) {
    return (x + y) % mod;
}

ll Sub(ll x, ll y) {
    return (x - y + mod) % mod;
}

ll res;
ll sum[N][2];

void dfs1(int u, int f) {
    sum[u][0] = sum[u][1] = 0;

    for (int i = 0, sz = e[u].size(); i < sz; i++) {
        int v = e[u][i].v;
        if (v != f) {
            dfs1(v, u);
            sum[u][0] += sum[v][0];
            sum[u][1] += sum[v][1];
        }
    }
    sum[u][a[u]]++;
}

void dfs2(int u, int f) {
    for (int i = 0, sz = e[u].size(); i < sz; i++) {
        int v = e[u][i].v;
        if (v != f) {
            dfs2(v, u);

            res = Add(res, Mul(e[u][i].w, Mul(sum[v][0], Sub(sum[1][1], sum[v][1]))));
            res = Add(res, Mul(e[u][i].w, Mul(sum[v][1], Sub(sum[1][0], sum[v][0]))));

            //res += sum[v][0] * (sum[1][1] - sum[v][1]);
            //res += sum[v][1] * (sum[1][0] - sum[v][0]);
        }
    }
}

ll p[N];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    p[0] = 1;
    for (int i = 1; i <= 200010; i++) {
        p[i] = Mul(p[i - 1], 2);
    }

    int T, u, v;
    cin >> T;
    while (T--) {
        cin >> n >> m;
        for (int i = 1; i <= n; i++) {
            cin >> a[i];

            // init
            e[i].clear();
            fa[i] = i;
        }

        for (int i = 1; i <= m; i++) {
            cin >> u >> v;
            Union(u, v, p[i]);
        }
        
        res = 0;
        dfs1(1, 0);
        dfs2(1, 0);

        cout << res << endl;
    }
    return 0;
}

A Very Easy Math Problem
Yukikaze and Smooth numbers

Divisibility

#include 
using namespace std;
typedef long long ll;
ll b, x;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int T;
    cin >> T;
    while (T--) {
        cin >> b >> x;
        if (b % x == 1) {
            cout << "T" << endl;
        } else {
            cout << "F" << endl;
        }
    }
    return 0;
}

Expectation
Kirakira