Codeforces_849

Posted 2020-10-05 冷暖知不知

A.只要考虑分成一个串的情况就可以了。

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

int n,a[105];

int main()
{
    ios::sync_with_stdio(0);
    cin >> n;
    for(int i = 1;i <= n;i++)   cin >> a[i];
    if(a[1]%2 && a[n]%2 && n%2)    cout << "Yes" << endl;
    else    cout << "No" << endl;
    return 0;
}

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B.1.若所有点在一条直线，则no。

2.若a[1]单独一组，剩余点在一条直线，则yes。

3.枚举a[1]和后面点组成的直线，判断是否能把点分成两组。

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

int n,a[1005];

int main()
{
    ios::sync_with_stdio(0);
    cin >> n;
    for(int i = 1;i <= n;i++)   cin >> a[i];
    int t = a[3]-a[2];
    int ok = 0;
    for(int i = 3;i <= n;i++)
    {
        if(a[i]-a[i-1] != t)    ok = 1;
    }
    if(!ok)
    {
        if(a[2]-a[1] == t)  cout << "No" << endl;
        else    cout << "Yes" << endl;
        return 0;
    }
    ok = 0;
    for(int i = 2;i <= n;i++)
    {
        double t = 1.0*(a[i]-a[1])/(i-1);
        vector<int> v;
        for(int j = 2;j <= n;j++)
        {
            if(abs(1.0*(a[j]-a[1])/(j-1)-t) > 1e-8) v.push_back(j);
        }
        int flag = 1;
        for(int j = 1;j < v.size();j++)
        {
            if(abs(1.0*(a[v[j]]-a[v[0]])/(v[j]-v[0])-t) > 1e-8)   flag = 0;
        }
        if(flag)    ok = 1;
    }
    if(ok)  cout << "Yes" << endl;
    else    cout << "No" << endl;
    return 0;
}

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C.首先，每个字母相互独立。对于每个字母，最优的方案是一个一个加，打个表，每次加个最大可行量。注意0的特殊情况。

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

int n,x[500];
string ans = "z";

int main()
{
    ios::sync_with_stdio(0);
    x[1] = 0;
    for(int i = 2;i <= 440;i++) x[i] = x[i-1]+i-1;
    cin >> n;
    for(char c = ‘a‘;c <= ‘z‘;c++)
    {
        if(n == 0)  break;
        int t = 440;
        while(x[t] > n) t--;
        n -= x[t];
        while(t--)  ans.append(1,c);
    }
    cout << ans << endl;
    return 0;
}

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D.分成两组，按照g-t排序，这个顺序保证相遇顺序。

之后双指针模拟，遇到连续的几个相同时，画个图便知道那个点从哪里出了。

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

int n,w,h,tx[100005],ty[100005],ansx[100005],ansy[100005];
struct xx
{
    int x,t,id;
    friend bool operator<(xx a,xx b)
    {
        if(a.x-a.t != b.x-b.t)  return a.x-a.t < b.x-b.t;
        return a.x < b.x;
    }
}a[100005],b[100005];

int main()
{
    ios::sync_with_stdio(0);
    cin >> n >> w >> h;
    int cnt1 = 0,cnt2 = 0;
    for(int i = 1;i <= n;i++)
    {
        int x,y,t;
        cin >> x >> y >> t;
        if(x == 1)
        {
            a[++cnt1].x = y;
            a[cnt1].t = t;
            a[cnt1].id = i;
            tx[i] = y;
            ty[i] = h;
        }
        else
        {
            b[++cnt2].x = y;
            b[cnt2].t = t;
            b[cnt2].id = i;
            tx[i] = w;
            ty[i] = y;
        }
    }
    sort(a+1,a+1+cnt1);
    sort(b+1,b+1+cnt2);
    int now1 = 1,now2 = 1;
    while(now1 <= cnt1 && now2 <= cnt2)
    {
        if(a[now1].x-a[now1].t == b[now2].x-b[now2].t)
        {
            int t1 = now1+1,t2 = now2+1;
            while(t1 <= cnt1 && a[now1].x-a[now1].t == a[t1].x-a[t1].t) t1++;
            while(t2 <= cnt2 && b[now2].x-b[now2].t == b[t2].x-b[t2].t) t2++;
            vector<int> v1,v2;
            for(int i = now1;i < t1;i++)    v1.push_back(a[i].id);
            for(int i = t2-1;i >= now2;i--) v1.push_back(b[i].id);
            for(int i = t2-1;i >= now2;i--) v2.push_back(b[i].id);
            for(int i = now1;i < t1;i++)    v2.push_back(a[i].id);
            for(int i = 0;i < v1.size();i++)
            {
                ansx[v2[i]] = tx[v1[i]];
                ansy[v2[i]] = ty[v1[i]];
            }
            now1 = t1;
            now2 = t2;
        }
        else if(a[now1].x-a[now1].t < b[now2].x-b[now2].t)
        {
            ansx[a[now1].id] = tx[a[now1].id];
            ansy[a[now1].id] = ty[a[now1].id];
            now1++;
        }
        else
        {
            ansx[b[now2].id] = tx[b[now2].id];
            ansy[b[now2].id] = ty[b[now2].id];
            now2++;
        }
    }
    while(now1 <= cnt1)
    {
        ansx[a[now1].id] = tx[a[now1].id];
        ansy[a[now1].id] = ty[a[now1].id];
        now1++;
    }
    while(now2 <= cnt2)
    {
        ansx[b[now2].id] = tx[b[now2].id];
        ansy[b[now2].id] = ty[b[now2].id];
        now2++;
    }
    for(int i = 1;i <= n;i++)   cout << ansx[i] << " " << ansy[i] <<endl;
    return 0;
}

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