标签：wap 主席树 operator char his 强制 dfs struct end
这可怜的分数
考场上强转的时候1ll写成了1l，60变28??
将最小包含自己的圆当作父亲，跑一个简单的树形DP就可以求出答案，考虑如何找父亲
扫描线，将扫到的圆的上下面分别加到set里，新加一个点的时候找y坐标第一个比它大的半圆，如果这个半圆是上半圆，那一定是它父亲，否则就是兄弟
我写这篇其实主要不是为了题解，而是关于using namespace std;
在郭神加强数据之后，一直TLE，看着比我写的还丑的HISKrrr都过了，调了一晚上，最后去掉std::，换上using namspace std;然后就过了
从今往后，如果再有人敢在我面前说写std::这样好那样好，我就当场把那个人的头咬下来！！！
#include <set>
#include <cmath>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 1e5 + 5;
int read(int x = 0, int f = 1, char c = getchar()) {
for (; c < ‘0‘ || c > ‘9‘; c = getchar())
if (c == ‘-‘) f = -1;
for (; c >=‘0‘ && c <=‘9‘; c = getchar())
x = x * 10 + c - ‘0‘;
return x * f;
}
struct Edge {
int n, t;
}e[N];
int h[N], edc;
void Add(int x, int y) {
e[++edc] = (Edge) {h[x], y}; h[x] = edc;
}
int n, x[N], y[N], r[N], w[N], fa[N], X, m;
struct Node {
int i, k;
double Cal() const {
return y[i] + k * sqrt(1ll * r[i] * r[i] - 1ll * (x[i] - X) * (x[i] - X));
}
};
set<Node> s;
set<Node>::iterator it;
bool operator < (const Node &a, const Node &b) {
double ya = a.Cal(), yb = b.Cal();
return ya != yb ? ya < yb : a.k < b.k;
}
struct Node2 {
int x, i, k;
}a[N*2];
bool operator < (const Node2 &a, const Node2 &b) {
return a.x < b.x;
}
int Dfs(int x) {
int s = 0;
for (int i = h[x]; i; i = e[i].n)
s += Dfs(e[i].t);
return s > w[x] ? s : w[x];
}
int main() {
n = read();
for (int i = 1; i <= n; ++i) {
x[i] = read(), y[i] = read(), r[i] = read(), w[i] = read();
a[++m] = (Node2) {x[i] - r[i], i, 1};
a[++m] = (Node2) {x[i] + r[i], i, 0};
}
sort(a + 1, a + m + 1);
for (int j = 1; X = a[j].x, j <= m; ++j) {
if (!a[j].k) s.erase((Node) {a[j].i, 1}), s.erase((Node) {a[j].i, -1});
else {
int z = a[j].i; it = s.lower_bound((Node) {z, 1});
if (it != s.end()) fa[z] = it->k == 1 ? it->i : fa[it->i];
s.insert((Node) {z, 1}), s.insert((Node) {z, -1});
}
}
for (int i = 1; i <= n; ++i) Add(fa[i], i);
printf("%d\n", Dfs(0));
return 0;
}
经过一系列推导，可得出ans(n,k)=lcm(n-k+1,n-k+2,…,n)
然后就是维护一个长为r的数组a，令
考虑已经求出长为r-1的数组，怎么维护长为r的数
其实就是把r质因数分解，然后在前面找到这个质因数在哪出现的，删去就好了
多组询问且强制在线的话写个主席树就好了
#include <cstdio>
#include <algorithm>
#define ls t[rt].l
#define rs t[rt].r
using namespace std;
const int N = 1e5 + 5, M = 1e9 + 7;
int read(int x = 0, int f = 1, char c = getchar()) {
for (; c < ‘0‘ || c > ‘9‘; c = getchar())
if (c == ‘-‘) f = -1;
for (; c >=‘0‘ && c <=‘9‘; c = getchar())
x = x * 10 + c - ‘0‘;
return x * f;
}
int q, n, k, a, b, m, c[N*2], d[N*2], ans;
int pri[N], tot, v[N], inv[N], s[N], rt[N], trc;
struct Tree {
int s, l, r;
Tree() { s = 1; }
}t[N*80];
void Add(int &rt, int l, int r, int x, int w) {
t[++trc] = t[rt]; rt = trc;
t[rt].s = 1ll * t[rt].s * w % M;
if (l == r) return;
int mid = l + r >> 1;
if (x <= mid) Add(ls, l, mid, x, w);
else Add(rs, mid + 1, r, x, w);
}
int Ask(int rt, int l, int r, int x, int y) {
if (x <= l && r <= y) return t[rt].s;
int mid = l + r >> 1, ans = 1;
if (x <= mid) ans = Ask(ls, l, mid, x, y);
if (y > mid) ans = 1ll * ans * Ask(rs, mid+1, r, x, y) % M;
return ans;
}
int main() {
q = read(); n = read(); k = read();
a = read(); b = read(); m = read();
for (int i = 1; i < q; ++i) c[i] = read();
for (int i = 1; i < q; ++i) d[i] = read();
inv[1] = 1;
for (int i = 2; i <= m; ++i) {
inv[i] = 1ll * (M - M / i) * inv[M%i] % M;
if (!v[i]) pri[++tot] = i, v[i] = i;
for (int j = 1; j <= tot && i * pri[j] <= m; ++j) {
v[i*pri[j]] = pri[j];
if (i % pri[j] == 0) break;
}
}
for (int i = 2; i <= m; ++i) {
int x = i, p; Add(rt[i] = rt[i-1], 1, m, i, i);
while (p = v[x]) {
int cnt = 0, mul = p;
while (x % p == 0) x /= p, cnt++;
if (cnt > s[p]) swap(cnt, s[p]);
while (cnt--) Add(rt[i], 1, m, i - mul, inv[p]), mul *= p;
}
}
for (int i = 1; i <= q; ++i) {
printf("%d\n", ans = Ask(rt[n], 1, m, n - k + 1, n));
n = (1ll * a * ans + c[i]) % m + 1;
k = (1ll * b * ans + d[i]) % n + 1;
}
return 0;
}
标签：wap 主席树 operator char his 强制 dfs struct end
原文地址：https://www.cnblogs.com/shawk/p/14290185.html
