第十一场训练赛

A - AAA

CodeForces - 34B

思路：

典型的贪心，在能搬动的情况下，如果值为负数就带走

#include<bits/stdc++.h>
using namespace std;
int main() {
	//freopen("in.txt", "r", stdin);
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	int n, m;cin >> n >> m;
	int a[100];
	for (int i = 0; i < n; ++i)cin >> a[i];
	sort(a, a + n);
	int cnt = 0;
	for (int i = 0; i < m; ++i)if (a[i] < 0)cnt += abs(a[i]);
	cout << cnt;
}

B - BBB

CodeForces - 1065B

思路：

数学问题，在满足能点构成的边会小于 (m) 时不断++ ,此时cnt是最大能连接的点

[Min：min(0,n - 2 * m)\ Max:n - cnt ]

#include<bits/stdc++.h> 
using namespace std;
typedef long long ll;
ll n, m, Min = 0, Max = 0;
int main() {
	//freopen("in.txt", "r", stdin);
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> m;
	ll cnt = 0;
	while (cnt * (cnt - 1) / 2 < m)cnt++;
	cout << max((ll)0, n - 2 * m) << " " << n - cnt << endl;
}

C - CCC

CodeForces - 1065C

思路：差分统计每个高度i有的方块数h[i]，然后从高到低贪心的切就行了

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 200000;
ll _, n, m, k, vis[N + 10];
int main() {
	//freopen("in.txt", "r", stdin);
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> k;
	for (int i = 0; i < n; i++) { int x; cin >> x; vis[x]++; }
	ll cnt = 0, t = 0;
	for (ll i = N; i > 0; --i) {
		if (vis[i] == n)break;
		if (vis[i] + t > k) {
			t = vis[i], cnt++;
		}
		else
			t += vis[i];
		vis[i - 1] += vis[i];
	}
	cout << (t ? cnt + 1 : cnt) << endl;
}

D - DDD

HDU - 6890

题意：

这个题把，应该算是思维题，最后出门的时候，带上最靠近门口的快递，其他的随便怎么拿，每拿一次就要回到K按一次密码

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll _, n, m, k;
string anw[2] = { "YES
","NO
" };
void solve() {
	cin >> n >> m >> k;
	//int a[m + 1];
	ll a[m + 1], sum = 0;
	for (int i = 1; i <= m; ++i) cin >> a[i];
	sort(a + 1, a + 1 + m);
	sum += k - 1;
	for (int i = m; i > 1; i--) {
		sum += 2 * abs(a[i] - k);
	}
	sum += abs(a[1] - k) + a[1] - 1;
	cout << sum << endl;
}
int main() {
	//freopen("in.txt", "r", stdin);
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> _; while (_--)solve();
}

E - EEE （签到题）

HDU - 6897

简单理解题意即可，签到

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll _, n;
string anw[2] = { "YES
","NO
" };
void solve() {
	cin >> n;
	int a[n + 1];
	for (int i = 1; i <= n; ++i)cin >> a[i];
	for (int i = 2; i <= n; ++i)if (a[i] == a[i - 1]) { cout << anw[1]; return; }
	cout << anw[0];
}
int main() {
	//freopen("in.txt", "r", stdin);
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> _; while (_--)solve();
}

F - 做出来算我输

HDU - 6889

涉及算法：Min25筛 1e10

题意：

n个点的完全图，边权为lcm(i+1,j+1)，求mst（最小生成树）

参考：https://blog.csdn.net/qq_39972971/article/details/81543972

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 1000010;
ll n, mod;
ll q_pow(ll a, ll b) {//快速取幂
    ll ans = 1;
    while (b){
        if (b & 1)
            ans = ans * a % mod;
        a = a * a % mod;
        b /= 2;
    }
    return ans % mod;
}
ll prime[N], id1[N], id2[N], flag[N], ncnt, m;
ll g[N], sum[N], a[N], T;
inline ll ID(ll x) {
	return x <= T ? id1[x] : id2[n / x];
}
inline ll calc(ll x) {
	return x * (x + 1) / 2 - 1;
}
inline ll f(ll x) {
	return x;
}
inline void init() {
    ncnt = m = 0;
    T = sqrt(n + 0.5);
    for (ll i = 2; i <= T; i++)
    {
        if (!flag[i])
            prime[++ncnt] = i, sum[ncnt] = sum[ncnt - 1] + i;
        for (ll j = 1; j <= ncnt && i * prime[j] <= T; j++)
        {
            flag[i * prime[j]] = 1;
            if (i % prime[j] == 0)
                break;
        }
    }
    for (ll l = 1; l <= n; l = n / (n / l) + 1)
    {
        a[++m] = n / l;
        if (a[m] <= T)
            id1[a[m]] = m;
        else
            id2[n / a[m]] = m;
        g[m] = calc(a[m]);
    }
    for (ll i = 1; i <= ncnt; i++)
        for (ll j = 1; j <= m && (ll)prime[i] * prime[i] <= a[j]; j++)
            g[j] = g[j] - (ll)prime[i] * (g[ID(a[j] / prime[i])] - sum[i - 1]);
}
inline ll solve(ll x) {
	if (x <= 1)return x;
	return n = x, init(), g[ID(n)];
}
int main() {
	//freopen("in.txt", "r", stdin);
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	int _; cin >> _;
	ll n;
	while (_--) {
		cin >> n >> mod;
		ll ans = (solve(n + 1) - 2 + mod) % mod;//质数和
		ll inv2 = q_pow(2, mod - 2) % mod;
		ll tmp = ((n + 4) % mod * (n - 1) % mod) % mod * inv2 % mod;
		cout << (ans + tmp) % mod << endl;
	}
}