2022 ICPC 杭州

唉我还是太菜了

A.Modulo Ruins the Legend

纠正一下：第一种情况最小是sum2 第二种情况最坏是sum2 可以比sum2更小

为什么？因为g是m的因数所以后面就会循环了

题解中说明：其实d只用取0或者1 因为相当于对每个位置加上一个平均数也就只加了ns 但是可能这个平均数可能为分数所以d取1和0即可代表所有情况

#include <bits/stdc++.h>
#define endl "\n"
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

ll exgcd(ll a, ll b, ll& x, ll& y) {
    if (!b) {
        x = 1, y = 0;
        return a;
    }
    ll d = exgcd(b, a % b, x, y);
    ll tx = x;
    x = y, y = tx - y * (a / b);
    return d;
}

signed main() {
    IOS;
    ll n, mod; cin >> n >> mod;
    ll sum = 0;
    for (int i = 1; i <= n; i++) {
        ll t; cin >> t;
        sum += t;
    }

    ll a = n, b = n * (n + 1) / 2;
    ll s, dt;
    ll d = exgcd(a, b, s, dt);
    sum %= mod;

    ll k, t;
    ll g = exgcd(d, mod, k, t);

    ll z = (mod - sum + g - 1) / g;
    (k *= z) %= mod;
    
    s = ((s % mod * k) % mod + mod) % mod, dt = ((dt % mod * k) % mod + mod) % mod;

    cout << (z * g + sum - mod) << endl;
    cout << s << " " << dt << endl;
}

C. No Bug No Game

分析：

选物品的顺序很重要但是题目并没有要求顺序

有一个关键的性质最多只有一个物品是取不完的

此时会有疑问当前转移到i 后面的不是还没转移嘛

其实并不是因为最多只有一个物品取不完后面的还是按照都选的转移即可

const int N = 3010;
long long dp[N][N][5];
void Immortality()
{
	int n, k;
	cin >> n >> k;
	long long sum = 0;
	vector<vector<int>> p(n + 1);
	for (int i = 1; i <= n; i++)
	{
		int t;
		cin >> t;
		sum += t;
		p[i].resize(t + 1);
		for (int j = 1; j <= t; j++)
			cin >> p[i][j];
	}
	if (sum <= k)
	{
		long long ans = 0;
		for (int i = 1; i <= n; i++)
		{
			int t = (p[i].size() - 1);
			ans += p[i][t];
		}
		cout << ans << endl;
		return;
	}
	memset(dp, -0x3f, sizeof dp);
	dp[0][0][1] = 0;
	dp[0][0][0] = 0;
	for (int i = 1; i <= n; i++)
	{
		int t = (p[i].size() - 1);
		for (int j = 0; j <= k; j++)
		{
			dp[i][j][0] = dp[i - 1][j][0];
			dp[i][j][1] = dp[i - 1][j][1];
			if (j >= t)
			{
				dp[i][j][0] = max(dp[i - 1][j][0], dp[i - 1][j - t][0] + p[i][t]);
				dp[i][j][1] = max(dp[i - 1][j][1], dp[i - 1][j - t][1] + p[i][t]);
			}
			for (int z = 1; z < t; z++)
				if (j >= z)
					dp[i][j][0] = max(dp[i][j][0], dp[i - 1][j - z][1] + p[i][z]);
		}
	}
	cout << max(dp[n][k][1], dp[n][k][0]) << endl;
}

K. Master of Both

题意：根据设置的字符大小顺序

求字符串逆序对的个数

分析：

很明显我们不能对每一种字符顺序求一次逆序对

#include "bits/stdc++.h"

using namespace std;
using i64 = long long;

constexpr int N = 26;

constexpr int M = 2E6;

i64 cnt[M + 10];
i64 add;
i64 num[N][N];
i64 ed[M + 10];

class Trie {
 public:
  int tot;
  vector<vector<int>> t;
  Trie(const int &n = 2E6) : t(n, vector<int>(N)), tot(0) {}
  inline void insert(const string &s) {
    int SIZE = s.size();
    int u = 0;
    for (int i = 0; i < SIZE; i++) {
      int c = s[i] - 'a';
      for (int j = 0; j < N; j++) {
        if (c != j && t[u][j]) {
          num[c][j] += cnt[t[u][j]];
        }
      }
      if (t[u][c] == 0) {
        t[u][c] = ++tot;
      }
      u = t[u][c];
      cnt[u]++;
    }
    ed[u]++;
    add += cnt[u] - ed[u];
  }
};

void solve() {
  int n, q;
  cin >> n >> q;
  Trie tree;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    tree.insert(s);
  }
  for (int i = 0; i < q; i++) {
    string s;
    cin >> s;
    i64 ans = 0;
    for (int j = 0; j < N; j++) {
      for (int k = j + 1; k < N; k++) {
        ans += num[s[j] - 'a'][s[k] - 'a'];
      }
    }
    cout << ans + add << '\n';
  }
}

int main() {
  int tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}