AtCoder Beginner Contest 132

AtCoder Beginner Contest 132

https://atcoder.jp/contests/abc132
持续被暴打的一天，因为晚上要打cf，所以明天再来写总结。悲
ct就是菜鸟newbie😫

A - Fifty-Fifty

#include <bits/stdc++.h>

using namespace std;

int main () {
    string s;
    cin >> s;
    map<char, int> mp;
    int cnt = 0;
    for (auto i : s)    mp[i]++;
    for (auto i : mp) {
        if (i.second != 2) {
            puts ("No");
            return 0;
        }
        cnt ++;
    }
    if (cnt != 2)   puts ("No");
    else    puts ("Yes");
}

B - Ordinary Number

#include <bits/stdc++.h>

using namespace std;
const int N = 25;
int a[N], n, ans;

int main () {
    cin >> n;
    for (int i = 1; i <= n; i++)    cin >> a[i];
    for (int i = 2; i < n; i++) {
        int maxn = max ({a[i], a[i-1], a[i+1]}), minn = min ({a[i], a[i-1], a[i+1]});
        if (a[i] != maxn && a[i] != minn)   ans ++;
    }
    cout << ans;
}

C - Divide the Problems

二分

#include <bits/stdc++.h>
#define int long long

using namespace std;
const int N = 1e5 + 5;
int a[N], n, ans, minn = 1e9, maxn;

signed main () {
    cin >> n;
    map<int, int> mp;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        minn = min (minn, a[i]), maxn = max (maxn, a[i]);
    }
    sort (a, a + n);
    for (int i = minn; i <= maxn; i++) {
        int pos = lower_bound (a, a + n, i) - a;
        //cout << i << ' ' << pos << endl;
        if (pos * 2 == n)   ans ++;
    }
    cout << ans;
}

D - Blue and Red Balls

组合计数

#include <bits/stdc++.h>
#define int long long

using namespace std;
const int N = 2005, p = 1e9 + 7;
int fact[N], infact[N], n, k;

int qmi(int a, int k) {
	int ans = 1;
	while (k) {
		if (k & 1)	ans = ans * a % p;
		k >>= 1;
		a = a * a % p;
	}
	return ans;
}

void init() {
    fact[0] = infact[0] = 1;
    for (int i = 1; i <= N - 3; i++) {
        fact[i] = fact[i - 1] * i % p;
        infact[i] = infact[i - 1] * qmi (i, p - 2) % p;
    }
}

int C(int a, int b) {
    if (a < b)      return 0;
    return fact[a] * infact[a - b] % p * infact[b] % p;
}

signed main () {
    init ();
    cin >> n >> k;
    for (int i = 1; i <= k; i++) {
        cout << (C(k - 1, i - 1) * C (n - k + 1, i)) % p << endl;
    }
}

E - Hopscotch Addict

建分层图。

#include <bits/stdc++.h>

using namespace std;
typedef pair<int, int> pii;
const int N = 1e5 + 5, M = N * 3;
int dis[M], n, m, st, ed;
int h[M], e[M], ne[M], idx;
bool vis[M];

void add (int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

int dijkstra () {
    priority_queue <pii, vector<pii>, greater<pii>>q;
    memset (dis, 0x3f, sizeof dis);
    dis[st] = 0;
    q.push ({0, st});
    
    while (!q.empty()) {
        auto t = q.top();
        q.pop();
        int ver = t.second, dist = t.first;
        
        if (vis[ver])  continue;
        vis[ver] = true;
        
        for (int i = h[ver]; i != -1; i = ne[i]) {
            int j = e[i];
            if (dis[j] > dist + 1) {
                dis[j] = dist + 1;
                q.push ({dis[j], j});
            }
        }
     }
    if (dis[ed] == 0x3f3f3f3f)    return -1;
    return dis[ed] / 3;
}

int main () {
    memset (h, -1, sizeof h);
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
        int a, b;
        cin >> a >> b; //建分层图
        add (a, b + n), add (a + n, b + n * 2), add (a + n * 2, b);
    }
    cin >> st >> ed;
    cout << dijkstra () << endl;
}

F - Small Products

整除分块优化dp
相似题目：https://www.luogu.com.cn/problem/P2261

#include <bits/stdc++.h>
#define ll long long

using namespace std;
const int M = 1e5 + 5, mod = 1e9 + 7;
ll n, k, m, f[M], tmp[M], a[M];

int main () {
    //cout << sqrt (1e9);
    cin >> n >> k;
    ll l = 1, r;
    while (l <= n) {
        if (n / l == 0)     r = n;
        else    r = n / (n / l);
        a[++m] = r - l + 1;
        l = r + 1;
    }
    for (int i = 1; i <= m; i++)    f[i] = 1;
    for (int i = 1; i <= k; i++) {
        for (int j = 1; j <= m; j++)    (tmp[j] = f[m-j+1] * a[j] % mod) %= mod;
        for (int j = 1; j <= m; j++)    (tmp[j] += tmp[j-1]) %= mod, f[j] = tmp[j];
    }
    cout << f[m] << endl;
}

//整除分块优化dp -> 根号n
//f[i][j]: i上放j, 滚掉一维