码迷，mamicode.com
首页 > 其他好文 > 详细

2020杭电多校第三场题解

时间：2020-07-29 14:30:59      阅读：162      评论：0      收藏：0      [点我收藏+]

标签：push   false   end   back   ++i   res   oid   否则   交点   

2020 Multi-University Training Contest 3

施工中。。。

1004 Tokitsukaze and Multiple

#include<bits/stdc++.h>
#define ll long long
#define maxn 100010
#define mod 998244353
using namespace std;
int a[maxn], sum[maxn], dp[maxn], las[maxn];
int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		int n, p;
		scanf("%d%d", &n, &p);
		int ans = 0;
		for (int i = 0; i <= p; i++) las[i] = 0;
		for (int i = 1; i <= n; i++) {
			int x;
			scanf("%d", &x);
			x %= p;
			a[i] = x;
		}
		for (int i = 1; i <= n; i++) sum[i] = (sum[i - 1] + a[i]) % p;
		for (int i = 1; i <= n; i++) {
			dp[i] = dp[i - 1];
			if (las[sum[i]] == 0 && sum[i] != 0) dp[i] = max(dp[i], dp[las[sum[i]]]);
			else dp[i] = max(dp[i], dp[las[sum[i]]] + 1);
			las[sum[i]] = i;
			ans = max(ans, dp[i]);
		}
		printf("%d\n", ans);
	}
	return 0;
}

1005 Little W and Contest

#include<bits/stdc++.h>
#define ll long long
#define maxn 100010
#define mod 1000000007
using namespace std;
ll f[maxn], s1[maxn], s2[maxn];
int fi(int x) {
	if (f[x] == x) return x;
	f[x] = fi(f[x]);
	return f[x];
}
ll po(ll x) {
	ll bas = 1, y = mod - 2;
	while (y) {
		if (y % 2) bas = bas * x % mod;
		x = x * x % mod;
		y /= 2;
	}
	return bas;
}
int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		int n;
		scanf("%d", &n);
		ll sum1 = 0, sum2 = 0;
		for (int i = 1; i <= n; i++) {
			int x;
			scanf("%d", &x);
			f[i] = i, s1[i] = s2[i] = 0;
			if (x == 1) s1[i]++, sum1++;
			else s2[i]++, sum2++;
		}
		ll ans = 0;
		ans = sum1 * (sum2 * (sum2 - 1) / 2) % mod % mod + (sum2 * (sum2 - 1) / 2 * (sum2 - 2) / 3) % mod;
		ans %= mod;
		vector<ll>v;
		v.push_back(ans);
		ll su2 = sum2 * (sum2 - 1) / 2 % mod;
		ll p2 = po(2), p6 = po(6);
		for (int i = 1; i < n; i++) {
			int x, y;
			scanf("%d%d", &x, &y);
			x = fi(x), y = fi(y);
			ans -= (sum1 - s1[x] - s1[y]) * s2[x] % mod * s2[y] % mod;
			ans %= mod;
			ans -= (sum2 - s2[x] - s2[y]) * s1[x] % mod * s2[y] % mod;
			ans %= mod;
			ans -= (sum2 - s2[x] - s2[y]) * s2[x] % mod * s1[y] % mod;
			ans %= mod;
			ans -= (sum2 - s2[x] - s2[y]) * s2[x] % mod * s2[y] % mod;
			ans %= mod;
			while (ans < 0) ans += mod;
			su2 -= s2[x] * s2[y] / 2;
			su2 %= mod;
			while (su2 < 0) su2 += mod;
			s1[x] += s1[y];
			s2[x] += s2[y];
			f[y] = x;
			v.push_back(ans);
		}
		for (int i = 0; i < v.size(); i++) printf("%lld\n", v[i]);
	}
	return 0;
}

1006 X Number

\(dp[pos][cnt][num]\) \(pos\) 为剩余的位数、\(cnt\) 为前面位数的组合情况、\(num\) 为要搜索的数的个数

本题要查询 \(0-9\) 每一位的位数，我们将 \(19\) 位分成若干个 \(0-9\) ，如\(\{(13,0),(2,1),(2,3),(1,4),(1,5),(0,6),(0,7),(0,8),(0,9)\}\)，大概有几千万种情况

事实上 \(0-9\) 对于我们来说只有等于 \(d\) 和不等于 \(d\) 的区别，那么仅把 \(19\) 个数分成若干块，如 \(\{13,2,2,1,1,0,0,0,0,0\}\)，只有约 \(2000\) 种情况

那么我们用 map<vector<int>, cnt> 来存储划分的情况，既可以在复杂度允许的情况，来表示划分。

接下来，用 \(num\) 表示要搜索的数的个数，唯一标识划分的情况

最后，数位 \(dp\) 暴力转移即可

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int maxn = 2005;
const int base = 25;

int Case = 0;

int g;
vector<int> tmp, E[maxn];
map<vector<int>, int> Map;
void init(int now, int k) {
    if (now == 10) {
        Map[tmp] = ++g;
        E[g] = tmp;
        return;
    }
    int top = k;
    if (now)
        top = min(top, tmp[now - 1]);
    for (int i = 0; i <= top; i++) {
        tmp.push_back(i);
        init(now + 1, k - i);
        tmp.pop_back();
    }
}

int a[base], len, d;
LL dp[base][maxn][base];

int limit_v[15], limit_cnt[base];
vector<int> limit_vc[base];

LL dfs(int pos, int cnt, int num, bool lead, bool limit) {
    if (pos > len) {
        if (E[cnt][0] > E[cnt][1] && E[cnt][0] == num) {
            return 1;
        } else
            return 0;
    }

    if (dp[len - pos + 1][cnt][num] != -1 && (!lead) && (!limit))
        return dp[len - pos + 1][cnt][num];

    LL res = 0;
    int top = limit ? a[len - pos + 1] : 9;

    if (lead) {
        res += dfs(pos + 1, cnt, num, lead, false);
        int top3 = top;
        if (limit)
            top3--;
        for (int i = 1; i <= top3; i++) {
            vector<int> tmp(10, 0);
            tmp[0] = 1;
            res += dfs(pos + 1, Map[tmp], num + (d == i), false, false);
        }
        if (limit)
            res += dfs(pos + 1, limit_cnt[pos], num + (d == top), false, true);
    } else if (limit) {
        vector<int> vc(10, 0);
        for (int i = 1; i < pos; i++)
            vc[a[len - i + 1]]++;
        for (int i = 0; i < top; i++) {
            vector<int> temp = vc;
            temp[i]++;
            sort(temp.begin(), temp.end(), [](const int a, const int b) {
                return a > b;
            });
            res += dfs(pos + 1, Map[temp], num + (d == i), false, false);
        }
        res += dfs(pos + 1, limit_cnt[pos], num + (d == top), false, true);
    } else {
        bool check = true;
        int add = 0;
        for (int i = 0; i < 10; i++) {
            vector<int> temp = E[cnt];
            if (check && temp[i] == num) {
                check = false;
                add = 1;
            } else
                add = 0;
            temp[i]++;
            sort(temp.begin(), temp.end(), [](const int a, const int b) {
                return a > b;
            });
            res += dfs(pos + 1, Map[temp], num + add, false, false);
        }
    }

    if ((!lead) && (!limit))
        dp[len - pos + 1][cnt][num] = res;
    return res;
}

LL doit(LL x) {
    len = 0;
    memset(a, 0, sizeof(a));
    while (x) {
        a[++len] = x % 10;
        x /= 10;
    }

    memset(limit_v, 0, sizeof(limit_v));
    for (int i = 1; i <= len; i++) {
        limit_v[a[len - i + 1]]++;
        limit_vc[i].clear();
        for (int j = 0; j <= 9; j++)
            limit_vc[i].push_back(limit_v[j]);
        sort(limit_vc[i].begin(), limit_vc[i].end(), [](const int a, const int b) {
            return a > b;
        });
        limit_cnt[i] = Map[limit_vc[i]];
    }

    return dfs(1, 1, 0, true, true);
}

int main() {
    init(0, 19);
    memset(dp, -1, sizeof(dp));

    int t;
    scanf("%d", &t);
    while (t--) {
        Case++;
        LL left, right;
        scanf("%lld%lld", &left, &right);
        scanf("%d", &d);
        printf("%lld\n", doit(right) - doit(left - 1));
    }
    return 0;
}

1007 Tokitsukaze and Rescue

#include<bits/stdc++.h>
#define ll long long
#define maxn 100010
#define mod 1000000007
using namespace std;
int n, k;
int mp[60][60];
int ans;
struct cv {
	int x, y;
	friend bool operator<(cv p, cv q) {
		return p.y > q.y;
	}
}dd, bb;
int las[60], len[60];
void sol(int x) {
	for (int i = 1; i <= n; i++) {
		las[i] = 0, len[i] = 1e8;
	}
	len[1] = 0;
	dd.x = 1, dd.y = 0;
	priority_queue<cv>q;
	q.push(dd);
	while (!q.empty()) {
		dd = q.top();
		q.pop();
		if (dd.x == n) break;
		if (dd.y != len[dd.x]) continue;
		for (int i = 1; i <= n; i++) {
			if (i == dd.x) continue;
			if (dd.y + mp[dd.x][i] < len[i]) {
				las[i] = dd.x;
				len[i] = dd.y + mp[dd.x][i];
				bb.x = i, bb.y = len[i];
				q.push(bb);
			}
		}
	}
	if (x == k) {
		ans = max(ans, len[n]);
		return;
	}
	int tp = n;
	vector<int>v;
	while (tp != 0) {
		v.push_back(tp);
		tp = las[tp];
	}
	for (int i = 0; i < v.size() - 1; i++) {
		int t1 = v[i], t3 = v[i + 1];
		int y = mp[t1][t3];
		mp[t1][t3] = mp[t3][t1] = 1e8;
		sol(x + 1);
		mp[t1][t3] = mp[t3][t1] = y;
	}
}
int main() {
	int  t;
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &n, &k);
		for (int i = 0; i < n * (n - 1) / 2; i++) {
			int x, y, z;
			scanf("%d%d%d", &x, &y, &z);
			mp[x][y] = mp[y][x] = z;
		}
		ans = 0;
		sol(0);
		printf("%d\n", ans);
	}
	return 0;
}

1008 Triangle Collision

二分时间。将正三角形展开，铺满整个平面，正三角形内的反射可以看作射线与展开后的正三角形平面的所有交点（如下图）。我们发现一共只有三种平行的直线，只要分别求出这条射线和这三种直线的交点分布就能 \(O(1)\) \(\text{check}\) 了。

技术图片
#include <bits/stdc++.h>
#define db double
using namespace std;
const db eps = 1e-5;
const db pi = acos(-1.0);

int sign(db k) {
	if (k > eps) return 1;
	else if (k < -eps) return -1;
	return 0;
}
int cmp(db k1, db k2) { return sign(k1 - k2); }

struct point {
	db x, y;
	point() {}
	point(db x_, db y_) :x(x_), y(y_) {}
	point operator + (const point& k) const { return point(k.x + x, k.y + y); }
	point operator - (const point& k) const { return point(x - k.x, y - k.y); }
	point operator * (db k) const { return point(x * k, y * k); }
	point operator / (db k1) const { return point(x / k1, y / k1); }
	point turn(db k1) { return point(x * cos(k1) - y * sin(k1), x * sin(k1) + y * cos(k1)); } // 逆时针旋转
	point turn90() { return point(-y, x); } // 逆时针方向旋转 90 度
	db len() { return sqrt(x * x + y * y); } // 向量长度
	db len2() { return x * x + y * y; } // 向量长度
	db dis(point rhs) { return ((*this) - rhs).len(); }
	point unit() { db d = len(); return point(x / d, y / d); }
	bool operator < (const point& k) const {
		return x == k.x ? y < k.y : x < k.x;
	}
	bool getP() const { return sign(y) == 1 || (sign(y) == 0 && sign(x) == -1); }
}triangleVertex[3];
db cross(point k1, point k2) { return k1.x * k2.y - k1.y * k2.x; }
db dot(point k1, point k2) { return k1.x * k2.x + k1.y * k2.y; }
db rad(point k1, point k2) { return atan2(cross(k1, k2), dot(k1, k2)); }
int compareangle(point k1, point k2) {
	return k1.getP() < k2.getP() || (k1.getP() == k2.getP() && sign(cross(k1, k2)) > 0);
}
point proj(point k1, point k2, point q) { // q 到直线 k1,k2 的投影
	point k = k2 - k1; return k1 + k * (dot(q - k1, k) / k.len2());
}
point reflect(point k1, point k2, point q) { return proj(k1, k2, q) * 2 - q; } // q 关于直线 k1,k2 的对称点
int clockwise(point k1, point k2, point k3) { // k1 k2 k3 逆时针1 顺时针-1 否则0
	return sign(cross(k2 - k1, k3 - k1));
}
int checkLL(point k1, point k2, point k3, point k4) { // 求直线(L) 线段(S) k1,k2 和 k3,k4 的交点
	return cmp(cross(k3 - k1, k4 - k1), cross(k3 - k2, k4 - k2)) != 0;
}

struct line {
	point p[2];
	line() {}
	line(point k1, point k2) { p[0] = k1, p[1] = k2; }
	point& operator [] (int k) { return p[k]; }
	point dir() { return p[1] - p[0]; }
	bool include(point k) { return sign(cross(p[1] - p[0], k - p[0])) > 0; }
	line push(db len) { // 向外（左手边）平移 len 个单位
		point delta = (p[1] - p[0]).turn90().unit() * len;
		return line(p[0] - delta, p[1] - delta);
	}
}triangle[3];

bool parallel(line k1, line k2) { return sign(cross(k1.dir(), k2.dir())) == 0; }
bool sameDir(line k1, line k2) { return parallel(k1, k2) && sign(dot(k1.dir(), k2.dir())) == 1; }
bool operator < (line k1, line k2) {
	if (sameDir(k1, k2)) return k2.include(k1[0]);
	return compareangle(k1.dir(), k2.dir());
}
point getLL(point k1, point k2, point k3, point k4) { // 两直线交点
	db w1 = cross(k1 - k3, k4 - k3), w2 = cross(k4 - k3, k2 - k3);
	return (k1 * w2 + k2 * w1) / (w1 + w2);
}
point getLL(line k1, line k2) { return getLL(k1[0], k1[1], k2[0], k2[1]); }
bool checkpos(line k1, line k2, line k3) { return k3.include(getLL(k1, k2)); }

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr); cout.tie(nullptr);
	int t; cin >> t;
	while (t--) {
		db len, x, y, vx, vy = 0.0;
		db L = 0.0, R = 1e20, mid;
		long long k;
		vector<db> initTime(3); // 第一次交直线的时间
		vector<db> crossTime(3); // 两次相交的间隔时间
		vector<bool> vis(3, false);
		cin >> len >> x >> y >> vx >> vy >> k;
		triangleVertex[0] = point(0.5 * len, 0);
		triangleVertex[1] = point(-0.5 * len, 0);
		triangleVertex[2] = point(0, 0.5 * sqrt(3.0) * len);
		triangle[0] = line(triangleVertex[0], triangleVertex[1]);
		triangle[1] = line(triangleVertex[1], triangleVertex[2]);
		triangle[2] = line(triangleVertex[2], triangleVertex[0]);
		point now = point(x, y);
		point dir = point(vx, vy) + now;
		line vec = line(now, dir);
		for (int i = 0; i < 3; ++i) {
			if (parallel(vec, triangle[i])) {
				vis[i] = true;
				continue;
			}
			point intersection = getLL(vec, triangle[i]);
			if (!sameDir(line(now, intersection), vec)) {
				triangle[i] = triangle[i].push(0.5 * sqrt(3.0) * len);
				intersection = getLL(vec, triangle[i]);
			}
			initTime[i] = now.dis(intersection);
			line nxtL = triangle[i].push(0.5 * sqrt(3.0) * len);
			point nxt = getLL(vec, nxtL);
			crossTime[i] = intersection.dis(nxt);
		}

		auto cal = [&](int i, db time) {
			if (vis[i]) return 0ll;
			long long cnt = 0;
			if (cmp(time, initTime[i]) == 1) {
				time -= initTime[i]; ++cnt;
				cnt += (long long)floor(time / crossTime[i]);
			}
			return cnt;
		};

		while (R - L > eps) {
			mid = 0.5 * (L + R);
			long long num = 0;
			for (int i = 0; i < 3; ++i) num += cal(i, mid);
			if (num >= k) R = mid;
			else L = mid;
		}
		cout << fixed << setprecision(10) << mid / sqrt(vx * vx + vy * vy) << ‘\n‘;
	}
	return 0;
}

1009 Parentheses Matching

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 5;
char s[maxn];

int main() {
	int t;
	scanf("%d", &t);

	while (t--) {
		scanf("%s", s + 1);
		int n = strlen(s + 1);
		stack<int>s1;
		queue<int>s2;
		for (int i = n; i > 0; i--) {
			if (s[i] == ‘)‘) s1.push(i);
			else if (s[i] == ‘(‘) {
				if (!s1.empty()) s1.pop();
				else s2.push(i);
			}
		}
		for (int i = 1; i <= n; i++) {
			if (s1.empty()) break;
			if (s[i] == ‘*‘) {
				int x = s1.top();
				if (i < x) {
					s[i] = ‘(‘;
					s1.pop();
				}
			}
		}
		for (int i = n; i >= 1; i--) {
			if (s2.empty()) break;
			if (s[i] == ‘*‘) {
				int x = s2.front();
				if (i > x) {
					s[i] = ‘)‘;
					s2.pop();
				}
			}
		}
		if ((!s1.empty()) || (!s2.empty())) {
			printf("No solution!\n");
		}
		else {
			for (int i = 1; i <= n; i++) {
				if (s[i] != ‘*‘) printf("%c", s[i]);
			}
			printf("\n");
		}
	}
	return 0;
}

2020杭电多校第三场题解

标签：push   false   end   back   ++i   res   oid   否则   交点   

原文地址：https://www.cnblogs.com/st1vdy/p/13396149.html
踩
(0)
赞
(0)
   
举报
评论 一句话评论（0）
登录后才能评论！
分享档案
更多>
2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)
周排行
mamicode.com排行更多图片
更多
友情链接
兰亭集智   国之画   百度统计   站长统计   阿里云   chrome插件   新版天听网
关于我们 - 联系我们 - 留言反馈
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码！