Codeforces_714

A.相遇时间段l = max(l1,l2),r = min(r1,r2)，再判断k是否在里面。

#include <iostream>
using namespace std;

long long l1,l2,r1,r2,k;

int main()
{
       cin >> l1 >> r1 >> l2 >> r2 >> k;
       long long l = max(l1,l2),r = min(r1,r2);
       if(l > r)    cout <<  << endl;
       else
       {
           long long t = r-l+;
           if(l <= k && k <= r)    t--;
           cout << t << endl;
    }
    return ;
}

---

B.判断出现的数的个数，1或2个直接YES，3个以上直接NO，3个判断是否等差。

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

int n,a[];
map<int,int> mp;

int main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    int cnt = ;
    for(int i = ;i <= n;i++)
    {
        int x;
        cin >> x;
        if(!mp.count(x))
        {
            cnt++;
            mp[x] = ;
            a[cnt] = x;
        }
    }
    if(cnt ==  || cnt == )    cout << "YES" << endl;
    else if(cnt >= )   cout << "NO" << endl;
    else
    {
        sort(a+,a+);
        if(a[]+a[] == *a[]) cout << "YES" << endl;
        else    cout << "NO" << endl;
    }
    return ;
}

---

C.把每一个数都转换成18位的pattern串，放进map处理。

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

int n;
map<string,int> mp;

int main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    while(n--)
    {
        string s1,s2,s = "";
        cin >> s1 >> s2;
        for(int i = s2.length()-;i >= ;i--)
        {
            if((s2[i]-'')%)   s = ""+s;
            else    s = ""+s;
        }
        while(s.length() < )  s = ""+s;
        if(s1 == "+")   mp[s]++;
        else if(s1 == "-")  mp[s]--;
        else    cout << mp[s] << endl;
    }
    return ;
}

---

D.先把图分成左右或上下两块，然后二分每一块中矩形的四条边。

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

int n;
struct xx
{
    long long x1,x2,y1,y2;
    void print()
    {
        cout << x1 << " " << y1 << " " << x2 << " " << y2 << " ";
    }
};

int query(int x1,int y1,int x2,int y2)
{
    if(x1 > x2)swap(x1,x2);
    if(y1 > y2)swap(y1,y2);
    cout<<"? "<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<endl;
    int ans;
    cin>>ans;
    return ans;
}

xx f(int x1,int y1,int x2,int y2)
{
    xx ans;
    long long l = x1,r = x2;
    while(l < r)
    {
        int mid = (l+r)/;
        if(query(x1,y1,mid,y2) < ) l = mid+;
        else    r = mid;
    }
    ans.x2 = l;
    l = x1,r = x2;
    while(l < r)
    {
        int mid = (l+r+)/;
        if(query(mid,y1,x2,y2) < ) r = mid-;
        else    l = mid;
    }
    ans.x1 = l;
    l = y1,r = y2;
    while(l < r)
    {
        int mid = (l+r)/;
        if(query(x1,y1,x2,mid) < ) l = mid+;
        else    r = mid;
    }
    ans.y2 = l;
    l = y1,r = y2;
    while(l < r)
    {
        int mid = (l+r+)/;
        if(query(x1,mid,x2,y2) < ) r = mid-;
        else    l = mid;
    }
    ans.y1 = l;
    return ans;
}
int main()
{
    cin >> n;
    long long x1,y1,x2,y2,x3,y3,x4,y4;
    long long l = ,r = n;
    while(l < r)
    {
        long long mid = (l+r)/;
        if(query(,,mid,n) < )    l = mid+;
        else    r = mid;
    }
    long long t = l;
    if(query(,,t,n) ==  && query(t+,,n,n) == )
    {
        xx a = f(,,t,n),b = f(t+,,n,n);
        cout << "! ";
        a.print();
        b.print();
        cout << endl;
        return ;
    }
    l = ,r = n;
    while(l < r)
    {
        long long mid = (l+r)/;
        if(query(,,n,mid) < )    l = mid+;
        else    r = mid;
    }
    t = l;
    xx a = f(,,n,t),b = f(,t+,n,n);
    cout << "! ";
    a.print();
    b.print();
    cout << endl;
    return ;
}

---

E.如果是非严格单调递增该如何做，我们会发现每次调整，都是调整某个数字为原先数列中存在的数字，最后才是最优的，所以，我们设DP[i][j]表示前i个数字，最后一个数为原先数列排序后第j大的数字的最小代价，然后令a[i]=a[i]-i，把严格单调递增就转化为非严格单调递增。

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

int n;
long long a[],b[],dp[][];

int main()
{
    cin >> n;
    for(int i = ;i <= n;i++)
    {
        cin >> a[i];
        a[i] -= i;
        b[i] = a[i];
    }
    sort(b+,b++n);
    for(int i = ;i <= n;i++)
    {
        long long minn = dp[i-][];
        for(int j = ;j <= n;j++)
        {
            minn = min(minn,dp[i-][j]);
            dp[i][j] = abs(a[i]-b[j])+minn;
        }
    }
    long long ans = dp[n][];
    for(int i = ;i <= n;i++)   ans = min(ans,dp[n][i]);
    cout << ans << endl;
    return ;
}

还有优先队列优化的。

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

int n;
priority_queue<long long> q;

int main()
{
    cin >> n;
    long long ans = ;
    for(int i = ;i <= n;i++)
    {
        long long x;
        cin >> x;
        x -= i;
        q.push(x);
        if(q.top() > x)
        {
            ans += q.top()-x;
            q.pop();
            q.push(x);
        }
    }
    cout << ans << endl;
    return ;
}