2023冲刺国赛模拟11

A. 天地一抹红

同行的转移是 \(kx + b\) 的形式，可以李超树维护

其实可以斜率优化，但是懒得动脑子

code

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, ll> pil;

int read(){
	int x = 0; char c = getchar();
	while(!isdigit(c))c = getchar();
	do{x = x * 10 + (c ^ 48); c = getchar();}while(isdigit(c));
	return x;
}
const int maxm = 2e4 + 55, maxn = 105;

ll calc(pil line, int x){return 1ll * line.first * x + line.second;}
int w[maxn][maxm], h[maxn][maxm], a[maxn][maxm];
int n, m, F;
struct seg{
	pil t[maxm << 2 | 1];
	void build(int x, int l, int r){
		t[x] = {0, -1e18};
		if(l == r)return;
		int mid = (l + r) >> 1;
		build(x << 1, l, mid);
		build(x << 1 | 1, mid + 1, r);
	}
	void modify(int x, int l, int r, pil line){
		int mid = (l + r) >> 1;
		if(calc(t[x], mid) < calc(line, mid))swap(t[x], line);
		if(l == r)return;
		if(calc(t[x], l) < calc(line, l))modify(x << 1, l, mid, line);
		if(calc(t[x], r) < calc(line, r))modify(x << 1 | 1, mid + 1, r, line);
	}
	ll query(int x, int l, int r, int v){
		ll ans = calc(t[x], v);
		if(l == r)return ans;
		int mid = (l + r) >> 1;
		if(v <= mid)return max(ans, query(x << 1, l, mid, v));
		else return max(ans, query(x << 1 | 1, mid + 1, r, v));
	}
}T;
ll f[maxn][maxm];
int main(){
	freopen("red.in","r",stdin);
	freopen("red.out","w",stdout);
	n = read(), m = read(), F = read();
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= m; ++j)
			w[i][j] = read();
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= m; ++j)
			h[i][j] = read();
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= m; ++j)
			a[i][j] = read();
    for(int i = 1; i <= n; ++i){
		T.build(1, 1, m); ll mx = -1e18;
		for(int j = 1; j <= m; ++j){
			if(i == 1 && j == 1)f[i][j] = F;
			else f[i][j] = max(f[i - 1][j] - w[i - 1][j], T.query(1, 1, m, a[i][j]));
			mx = max(f[i][j] - w[i][j], mx);
			T.modify(1, 1, m, pil(h[i][j], mx));
		}
	}
	printf("%lld\n", max(-1ll, f[n][m]));
	return 0;
}

B. 最简根式

不难发现是求 \(\sum_{a = 1}^n\sum_{b = 1}^{n}[\gcd(a, b) 含平方因子]\)

枚举 \(gcd\) 的平方因子，简单容斥得到答案为

\(-\sum_{i = 2}^{\sqrt n} \mu(i)(\lfloor\frac{n}{i^2}\rfloor)^2\)

对 \(\lfloor\frac{n}{i^2}\rfloor\) 整除分块

预处理一部分，杜教筛一部分，复杂度是对的

code

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, ll> pil;

ll read() {
    ll x = 0;
    char c = getchar();
    while (!isdigit(c)) c = getchar();
    do {
        x = x * 10 + (c ^ 48);
        c = getchar();
    } while (isdigit(c));
    return x;
}
const int mod = 998244353;
const int maxn = 6e7 + 5;
int prime[maxn], cnt, mu[maxn];
bool flag[maxn];
void init(int n) {
    mu[1] = 1;
    for (int i = 2; i <= n; ++i) {
        if (!flag[i])
            prime[++cnt] = i, mu[i] = -1;
        for (int j = 1; j <= cnt && prime[j] * i <= n; ++j) {
            flag[i * prime[j]] = true;
            if (i % prime[j] == 0)
                break;
            mu[i * prime[j]] = -mu[i];
        }
    }
    for (int i = 2; i <= n; ++i) mu[i] += mu[i - 1];
}
unordered_map<ll, int> mp;
int calc(ll n) {
    if (n <= 6e7)
        return mu[n];
    if (mp.count(n))
        return mp[n];
    ll ans = 1;
    for (ll l = 2, r; l <= n; l = r + 1) {
        r = n / (n / l);
        ans = (ans - 1ll * (r - l + 1) % mod * calc(n / l)) % mod;
    }
    ans = (ans + mod) % mod;
    return mp[n] = ans;
}
void solve() {
    ll n = read(), ans = 0;
    for (ll l = 2, r; l * l <= n; l = r + 1) {
        r = sqrt(n / (n / (l * l)));
        ans = (ans +
               1ll * ((n / (l * l)) % mod) * ((n / (l * l)) % mod) % mod * (calc(r) - calc(l - 1)) % mod) %
              mod;
    }
    printf("%lld\n", ((mod - ans) % mod + mod) % mod);
}
int main() {
    freopen("math.in", "r", stdin);
    freopen("math.out", "w", stdout);
    init(6e7);
    int t = read();
    for (int i = 1; i <= t; ++i) solve();
    return 0;
}

C. 图床

\(hash\) 找出出现的字符串个数 \(k\)

\[ P(n) = \frac{\binom{n}{k}(k^m - (k - 1)^m)}{n^m} \]

找到最接近的 \(n\)

这是个单峰函数

又有 \(P(n) < P(n + 1) <=> (n + 1 - k)(n + 1)^m < (n + 1)n^m\)

可以直接二分

code

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
const int mod = 2140207;
int read() {
    int x = 0;
    char c = getchar();
    while (c < 'a' || c > 'z') c = getchar();
    do {
        x = (x * 29ll + (c - 'a' + 1)) % mod;
        c = getchar();
    } while (c >= 'a' && c <= 'z');
    return x;
}
int ln, rn, m;
bitset<mod> rem;
void solve() {
    int k = 0;
    for (int i = 1; i <= m; ++i) {
        int tmp = read();
        if (!rem[tmp])
            rem[tmp] = true, ++k;
    }
    int l = max(ln, k), r = rn;
    while (l < r) {
        int mid = (l + r) >> 1;
        if (log2(mid - k + 1) + log2(mid + 1) * m < log2(mid + 1) + log2(mid) * m)
            l = mid + 1;
        else
            r = mid;
    }
    rem.reset();
    printf("%d\n", l);
}

int main() {
    freopen("pic.in", "r", stdin);
    freopen("pic.out", "w", stdout);
    int T;
    double a, b, c, d;
    scanf("%d%lf%lf%lf%lf%d%d%d", &T, &a, &b, &c, &d, &ln, &rn, &m);
    for (int i = 1; i <= T; ++i) solve();
    return 0;
}