2022icpc杭州铜牌题题解

A. Modulo Ruins the Legend

$$求s、d，使\sum a_i +sn+d\frac{n(n+1)}{2} \ (\bmod m)最小\\ 设sum = \sum a_i\ (\bmod m)，t=gcd(n,\frac{n(n+1)}{2})\\ 原式=sum+kt\ (\bmod m)\\ 这时还不能求最小值，因为t和m没啥关系，可以添项\\ sum+kt\ (\bmod m) \Leftrightarrow sum+kt+pm\ (\bmod m)\\ 再做一步扩欧，设g=gcd(t,m)\\ 原式=sum+qg\ (\bmod m)\\ g\ |\ m，循环节的长度是m/g\\ 当q=\left \lceil \frac{m-sum}{g} \right \rceil时原式最小$$

LL exgcd(LL a, LL b, LL &x, LL &y) {
    if (!b) {
        x = 1, y = 0;
        return a;
    }
    LL d = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}
 
void work() { 
	LL n, m;
	cin >> n >> m; 
	
	LL sum = 0;
	rep (i, 1, n) {
		LL x;
		cin >> x;
		sum = (sum + x) % m;
	}
	
	LL x_s, y_d;
	LL t = exgcd(n, n * (n + 1) / 2, x_s, y_d);
	
	LL x_k, y_p;
	LL g = exgcd(t, m, x_k, y_p);
	
	LL q = (m - sum + g - 1) / g;
	
	LL s = q * x_k % m * x_s % m, d = q * x_k % m * y_d % m;
	cout << (q * g + sum) % m << endl;
	cout << (s + m) % m << " " << (d + m) % m << endl;
}

C. No Bug No Game

$$状态表示：dp[i][j][0/1]，表示从前i个物品中选，体积为j，是否完全选择的最大值\\ 状态转移：分别为第i个物品不选，完全选择第i个物品，不完全选择第i个物品。因为不完全选择至多一次，所以不完全只能从完全转移。\\ 初始状态:dp[i][j][0/1] = -inf,dp[0][0][0]=0\\ $$

int n, k, s;
int p[N][11], dp[N][N][2]; // dp[i][j][0/1]：从前i个物品中选，体积为j，是否完全的最大值 
 
void work() { 
	cin >> n >> k;
	rep (i, 1, n) {
		cin >> p[i][0];
		s += p[i][0];
		rep (j, 1, p[i][0]) cin >> p[i][j];
	}
	
	memset(dp, -0x3f, sizeof dp);
	dp[0][0][0] = 0;
	rep (i, 1, n) {
		rep (j, 0, k) {
			//不选第i个物品
			dp[i][j][0] = dp[i - 1][j][0];
			dp[i][j][1] = dp[i - 1][j][1];
			
			//完全选择第i个物品 
			if (p[i][0] <= j) {
				dp[i][j][0] = max(dp[i][j][0], dp[i - 1][j - p[i][0]][0] + p[i][p[i][0]]); //完全更新完全
				dp[i][j][1] = max(dp[i][j][1], dp[i - 1][j - p[i][0]][1] + p[i][p[i][0]]); //完全更新不完全 
			}
			
			//不完全选择第i个物品 
			rep (t, 1, p[i][0] - 1) {
				if (t <= j) dp[i][j][1] = max(dp[i][j][1], dp[i - 1][j - t][0] + p[i][t]); //不完全更新完全 
				else break;
			}
		}
	}
	
	if (s <= k) cout << dp[n][s][0] << endl;
	else cout << max(dp[n][k][0], dp[n][k][1]) << endl; 
}

K. Master of Both

$$不论字母表怎么变，比较两个字符串的相应字母是不会变的，可以用字典树把所有字母对个数预处理出来，在字母表上求出对应的贡献即可$$

#define int long long
 
int n, q;
int son[N][30], cnt[N][30], idx = 1;
int mp[26][26], ex;
 
void insert(string s){
    int cur = 0;
    for (int i = 0; i < s.length(); i++) {
        if (!son[cur][s[i] - 'a']) son[cur][s[i] - 'a'] = idx++;
		rep (j, 0, 25) {
			if (j == s[i] - 'a') continue;
			mp[s[i] - 'a'][j] += cnt[cur][j]; 
		}
		cnt[cur][s[i] - 'a']++;
        cur = son[cur][s[i] - 'a'];
    }
	rep (i, 0, 25) ex += cnt[cur][i]; 
}
 
void work() {
	cin >> n >> q;
	rep (i, 1, n) {
		string s;
		cin >> s;
		insert(s);
	}
	rep (i, 1, q) {
		string s;
		cin >> s;
		int res = ex;
		rep (i, 0, 25) rep (j, i + 1, 25) res += mp[s[i] - 'a'][s[j] - 'a'];
		cout << res << endl;
	}
}