【记录】绵东实三校联考 验题记录

感觉已经快成御用验题小天使 = =。。

姜神出的题，然后cyx小朋友成功AK了

 

【T1】 排队 Waiting

这题和之前有一道NOIp题差不多。。做法也是一样的。。排个序就万事大吉了。

听说有人最后计算结果的时候少算了b，然后出了点儿事。。。

 

#include <cstdio>#include <iostream>#include <algorithm>#define Mod 1000000009llusing namespace std;int N, a[10005], b[10005], w[10005];bool cmp(const int &i, const int &j) { return a[i] * b[j] < a[j] * b[i]; };int main(){freopen("waiting.in", "r", stdin);freopen("waiting.out", "w", stdout);ios :: sync_with_stdio(false);cin >> N;for (int i = 1; i <= N; ++ i) {cin >> a[i] >> b[i]; w[i] = i;}sort(w + 1, w + N + 1, cmp);long long ans = 0ll;for (int i = 1; i <= N; ++ i) {ans += ans * b[w[i]] + a[w[i]]; if (ans > Mod) ans %= Mod; }cout << ans << endl;return 0;}

 

【T2】树上未解之谜

 

毕竟这玩意儿是我意淫出来的。。。但是改成二叉树君就太水了……直接把N^2变成了loglog

于是乎暴力玩玩就出来了……姜神的做法好像很诡异啊而且效率还没暴力高…………

 

#include <cstdio>#include <iostream>#include <cmath>using namespace std;const long long mod = 1000000009;int N, M, w[20005], high;int sum[20005], num[20005][20], add[20];#define lc (u << 1)#define rc (u << 1 | 1)void dfs(int u, int dep){if (u > N) return;++ num[u][dep]; sum[u] = w[u];dfs(lc, dep + 1); dfs(rc, dep + 1); sum[u] += sum[lc] + sum[rc];for (int i = dep + 1; i <= high; ++ i) num[u][i] += num[lc][i] + num[rc][i];}int main(){freopen("treeplus.in", "r", stdin);freopen("treeplus.out", "w", stdout);ios :: sync_with_stdio(false);cin >> N >> M; high = (int) (log(N) / log(2)) + 1;for (int i = 1; i <= N; ++ i) cin >> w[i];dfs(1, 1);int sign, a, b;while (M --) {cin >> sign;if (sign == 1) {cin >> a >> b; add[a] += b;} else {cin >> a; long long ans = sum[a];for (int i = 1; i <= high; ++ i) {ans += (long long)add[i] * num[a][i];if (ans > mod) ans %= mod;}cout << ans << endl;}}return 0;}

 

【T3】线段覆盖

 

姜神最开始的数据出了点问题……于是就一直拍不上…………大早上起来改数据心塞……

两棵线段树就可以搞定了……不知道能不能合到一起去。

 

#include <cstdio>#include <iostream>#include <algorithm>using namespace std;int N, M, l[1000005], r[1000005], bfr[2000005];#define lowbit(i) (i & -i)namespace BIT {int d[2000005][2];inline void Add(int pos, int c){for (int i = pos; i <= 2000000; i += lowbit(i)) ++ d[i][c];}inline int Query(int pos, int c){int ans = 0;for (int i = pos; i; i -= lowbit(i)) ans += d[i][c];return ans;}}int main(){freopen("segment.in", "r", stdin);freopen("segment.out", "w", stdout);ios :: sync_with_stdio(false);cin >> N; int L, R;for (int i = 1; i <= N; ++ i) {cin >> L; R = L + i - 1; cout << BIT :: Query(R, 1) - BIT :: Query(L - 1, 0) << endl; BIT :: Add(L, 0); BIT :: Add(R, 1);}return 0;}

 

【T4】魔法门

 

人生的第二道计算几何……调了好久居然是因为数组炸了。。

为什么溢出不会RE而会直接覆盖。。。

写了两个版本有一个会超时……另一个好像效率还蛮高的…………

还是觉得几何这种东西好烦TAT

 

然后最后这题被砍了……因为超纲了……不过练代码能力还是ok的……

 

#include <cstdio>#include <iostream>#include <cmath>#include <queue>#define eps 1e-15 using namespace std;int N;struct Point {double x, y;}P[5050];struct Vector {Point s, e;}wall[1050], Line[505 * 5050];int cntw, cntl, cntp;struct LinePoint {int sp, ep;}lp[505 * 5050];Vector SetLine(const double &sx, const double &sy, const double &ex, const double &ey){Point ss = (Point) { sx, sy };Point ee = (Point) { ex, ey };return (Vector) { ss, ee };}inline double Cross(Point A, Point B, Point O)  {// OA × OB     return (A.x - O.x) * (B.y - O.y) - (B.x - O.x) * (A.y - O.y);    }  bool intersect(Vector u, Vector v){// 判断线段相交 return ( Cross(v.s, u.e, u.s) * Cross(u.e, v.e, u.s) > eps &&   Cross(u.s, v.e, v.s) * Cross(v.e, u.e, v.s) > eps );}inline double Distance(double x1, double y1, double x2, double y2){return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));}namespace spfa {struct ed {int v, next;double w;}e[505 * 505 * 20];int head[5050], k = 1;inline void adde(int u, int v, double w){e[k] = (ed) {v, head[u], w};head[u] = k ++;}queue <int> q;double dis[5050];bool vis[5050];double SPFA(int S, int T){for (int i = 1; i <= cntp; ++ i) dis[i] = 1e100;dis[S] = 0. ; q.push(S);while (q.size()) {int u = q.front(); q.pop(); vis[u] = 0;for (int i = head[u]; i; i = e[i].next) {int v = e[i].v; if (dis[v] - eps > dis[u] + e[i].w) {dis[v] = dis[u] + e[i].w;if (!vis[v]) { vis[v] = 1; q.push(v); }}}}return dis[T];}}int main(){freopen("door.in", "r", stdin);freopen("door.out", "w", stdout);cin >> N; double x, a, b, c, d;P[++ cntp] = (Point) { 0. , 5. }; int S = cntp;for (int i = 1; i <= N; ++ i) {cin >> x >> a >> b >> c >> d;wall[++ cntw] = SetLine(x, 0, x, a);wall[++ cntw] = SetLine(x, b, x, c);wall[++ cntw] = SetLine(x, d, x, 10);P[++ cntp] = (Point) { x, a };P[++ cntp] = (Point) { x, b };P[++ cntp] = (Point) { x, c };P[++ cntp] = (Point) { x, d };}P[++ cntp] = (Point) { 10. , 5. }; int T = cntp;for (int i = 1; i <= cntp; ++ i) {for (int j = 1; j <= cntp; ++ j) {Line[++ cntl] = SetLine(P[i].x, P[i].y, P[j].x, P[j].y);lp[cntl] = (LinePoint) { i, j };}}for (int i = 1; i <= cntl; ++ i) {bool flag = 1; if (lp[i].sp == 2) {int a = 1;}for (int j = 1; j <= cntw; ++ j) if (intersect(Line[i], wall[j])) { flag = 0; break;}if (! flag) continue;int u = lp[i].sp, v = lp[i].ep; double w = Distance(Line[i].s.x, Line[i].s.y, Line[i].e.x, Line[i].e.y);spfa :: adde(u, v, w);  }  printf("%.2f\n", spfa :: SPFA(S, T));return 0;}