gym101522 [小熊骑士限定]La Salle-Pui Ching Programming Challenge 培正喇沙編程挑戰賽 2017

西瓜队（划掉），Kuma Rider久违的第一场训练，四小时瞎打.jpg

A.水题，排序

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
#include<string>
#include<vector>
#include<map>
#include<functional>
#include<vector>
using namespace std;
typedef long long ll;
#define lowbit(i) ((i)&(-i))
const ll maxv = 1e5 + ;

ll n,d[maxv], c[maxv];

void update(ll x, ll v)
{
    for (ll i = x; i < maxv; i+=lowbit(i))
    {
        c[i] += v;
    }
}

ll getsum(ll x)
{
    ll sum = ;
    for (ll i = x; i > ; i -= lowbit(i))
    {
        sum += c[i];
    }
    return sum;
}

int main()
{
    scanf("%lld", &n);
    for (ll i = ; i <= n; i++)
    {
        scanf("%lld", &d[i]);
    }
    sort(d+ , d+ + n);
    ll ans = ;
    for (ll i = ; i <=  n; i++)
    {
        ll num = getsum(i-);
        num = (i - ) - num;
        //printf("%lld %lld\n", i, num);
        ans += num;
        update(d[i], );
    }
    ans *= ;
    printf("%lld\n", ans);
}

B.找到树的直径，倍增地跳跃计算答案

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
#include<string>
#include<vector>
#include<map>
#include<functional>
#include<vector>
using namespace std;
typedef long long LL;

int n;
vector<int>e[];
bool vis[];
int Root, Tail;

int BFS(int st)
{
    for (int i = ; i <= n; i++)vis[i] = false;
    queue<int>q;
    q.push(st);
    vis[st] = true;
    int ret = st;
    while (!q.empty())
    {
        int t = q.front(); q.pop();
        ret = t;
        for (int i = ; i < e[t].size(); i++)
        {
            int to = e[t][i];
            if (!vis[to])
            {
                q.push(to);
                vis[to] = true;
            }
        }
    }
    return ret;
}

int len;

void DFS(int nx, int pr, int st)
{
    len = max(len, st);
    for (int i = ; i < e[nx].size(); i++)
    {
        int to = e[nx][i];
        if (to != pr)
        {
            DFS(to, nx, st + );
        }
    }
}

int main()
{
    scanf("%d", &n);
    for (int i = ; i <= n - ; i++)
    {
        int u, v;
        scanf("%d %d", &u, &v);
        e[u].push_back(v);
        e[v].push_back(u);
    }
    Root = BFS();
    Tail = BFS(Root);
    len = ;
    DFS(Root, -, );
    //printf("%d  %d\n", Root, Tail);
    int ans = ;
    int tl = ;
    while (tl<len-)
    {
        ans++;
        tl <<= ;
    }
    printf("%d\n", ans);

}

C.水题

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
#include<string>
#include<vector>
#include<map>
using namespace std;

char s[];
int main()
{
    scanf("%s", s);
    int len = strlen(s);
    int lsc = , pcms = ;
    for (int i = ; i < len-; i++)
    {
        if (s[i] == 'L'&&s[i + ] == 'S'&&s[i + ] == 'C')lsc++;
    }
    for (int i = ; i < len - ; i++)
    {
        if (s[i] == 'P'&&s[i + ] == 'C'&&s[i + ] == 'M'&&s[i + ] == 'S')pcms++;
    }
    if (pcms > lsc)printf("PCMS\n");
    else if (pcms < lsc)printf("LSC\n");
    else printf("Tie\n");

}

D.比赛没来得及写的题。（1953.1.1居然是星期六！！一直以为是星期日！！！）

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
#include<map>
#include<vector>
#include<set>
#include<queue>
using namespace std;
int days[] = { , ,  , , , , , , , , , , };

int q, st, ed, day, month, gi[], ans[];

bool isleap(int year)
{
    return (year %  == ) || ((year %  == ) && (year % ));
}

int cal(int yy, int mm, int dd)
{
    int sum = ;
    for (int i = ; i < mm; i++)
    {
        sum += days[i];
        if (isleap(yy) && i == )
            sum++;
    }
    return sum+dd-;
}

void cycle(int mm,int dd)
{
    int cnt=,now = ;//1953/1/1是星期六
    for (int i = ; i <=  + ; i++)
    {
        if (mm ==  && dd ==  && !isleap(i))
        {
            ans[cnt++] = ;
        }
        else
        {
            ans[cnt++] = (cal(i, mm, dd) + now - ) %  + ;
        }
        if (isleap(i))
            now = (now +  - ) %  + ;
        else
            now = (now +  - ) %  + ;
    }
}

int main()
{
    scanf("%d", &q);
    while (q--)
    {
        memset(ans, , sizeof(ans));
        memset(gi, , sizeof(gi));
        scanf("%d %d %d %d", &st, &ed, &month, &day);
        cycle(month, day);
        int numed = (ed- + ) / ;
        int numst = (st - ) / ;
        for (int i = ; i <= ; i++)
        {
            gi[ans[i]] += (numed - numst);
        }
        int mped = ed -  +  - numed * ;
        int mpst = st -  - numst * ;
        for (int i = ; i <= mpst; i++)
        {
            gi[ans[i]]--;
        }
        for (int i = ; i <= mped; i++)
        {
            gi[ans[i]]++;
        }
        printf("%d", gi[]);
        for (int i = ; i <= ; i++)
        {
            printf(" %d", gi[i]);
        }
        puts("");
    }
    return ;
}

E.还没看，待补

F.还没写，待补

G.还没写，待补

H.几何板子题，两圆交点，我抄错板子错了好久quq

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
#include<string>
#include<vector>
#include<cmath>
#include<map>
#include<functional>
#include<vector>
using namespace std;
const double eps = 1e-;

struct Point
{
    double x, y;
    Point(double _a, double _b) :x(_a), y(_b) {}
    Point() {}
};

Point intersection(const Point &u1, const Point &u2, const Point &v1, const Point &v2)
{
    Point res = u1;
    double t = ((u1.x - v1.x)*(v1.y - v2.y) - (u1.y - v1.y)*(v1.x - v2.x)) / ((u1.x - u2.x)*(v1.y - v2.y) - (u1.y - u2.y)*(v1.x - v2.x));
    res.x += (u2.x - u1.x)*t;
    res.y += (u2.y - u1.y)*t;
    return res;
}

double dis(const Point &p1, const Point &p2)
{
    return sqrt((p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y));
}

void play(const Point &c, double r, const Point &l1, const Point &l2, Point &p1, Point &p2)
{
    Point p = c;
    p.x += l1.y - l2.y;
    p.y += l2.x - l1.x;
    p = intersection(p, c, l1, l2);
    double t = sqrt(r*r - dis(p, c)*dis(p, c)) / dis(l1, l2);
    p1.x = p.x + (l2.x - l1.x)*t;
    p1.y = p.y + (l2.y - l1.y)*t;
    p2.x = p.x - (l2.x - l1.x)*t;
    p2.y = p.y - (l2.y - l1.y)*t;
}

void play2(const Point &c1, double r1, const Point &c2, double r2, Point &p1, Point &p2)
{
    Point u, v;
    double t = ( + (r1*r1 - r2*r2) / dis(c1, c2) / dis(c1, c2)) / ;
    u.x = c1.x + (c2.x - c1.x)*t;
    u.y = c1.y + (c2.y - c1.y)*t;
    v.x = u.x + c1.y - c2.y;
    v.y = u.y - c1.x + c2.x;
    play(c1, r1, u, v, p1, p2);
}

double x1, x2, r1, yy, y2, r2;

int main()
{
    cin >> x1 >> yy >> r1;
    cin >> x2 >> y2 >> r2;
    Point ans1, ans2;
    if (r1 > r2) {
        swap(x1, x2);
        swap(yy, y2);
        swap(r1, r2);
    }
    double ds = dis(Point(x1, yy), Point(x2, y2));
    if (r2 >= ds)
    {
        printf("%.6lf %.6lf\n", x1, yy);
        return ;
    }
    play2(Point(x1, yy), r1, Point(x2, y2), r2, ans1, ans2);
    printf("%.6lf %.6lf\n", ans1.x, ans1.y);
    return ;
}

I.瞎搞

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
#include<string>
#include<vector>
#include<cmath>
#include<map>
#include<functional>
#include<vector>
using namespace std;
typedef long long ll;
const double eps = 1e-;
ll ax[];
int main(void) {
    ll n;
    scanf("%lld", &n);
    for (int i = ; i < n; i++) {
        scanf("%lld", &ax[i]);
    }
    if (n == ) {
        printf("0\n");
        return ;
    }
    else if (n == ) {
        printf("%lld\n", min(abs(ax[] - ax[]),abs(ax[]+ax[])));
        return ;
    }
    vector<ll>v;
    ll sum = ;
    for (int i = ; i<n - ; i++) {
        sum += abs(ax[i + ] - ax[i]);
        v.push_back(abs(ax[i + ] + ax[i]) - abs(ax[i + ] - ax[i]));
    }
    sort(v.begin(), v.end());
    if (v[] < ) {
        sum += v[];
    }
    if (v[] < ) {
        sum += v[];
    }
    printf("%lld\n", sum);
}

J.还没写，待补

K.水题，四个角满足即可

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
#include<string>
#include<vector>
#include<map>
#include<functional>
#include<vector>
using namespace std;
typedef long long LL;

int n;

int main()
{
    int n, m, k;
    int cnt = ;
    scanf("%d %d %d", &n, &m, &k);
    for (int i = ; i <= k; i++)
    {
        int x, y;
        scanf("%d %d", &x, &y);
        if (x ==  && y == m)
            cnt++;
        else if (x ==  && y == )
            cnt++;
        else if (x == n&&y == )
            cnt++;
        else if (x == n&&y == m)
            cnt++;
    }
    int tmp = ;
    if (n ==  && m == )
        tmp = ;
    else if (n ==  || m == )
        tmp = ;
    printf("%d\n", tmp - cnt);
    return ;
}

L.还没写，待补