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牛客周赛 Round 10

A

O(n)一扫就行

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#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
 
int main()
{
    int n;
    cin >> n;
    vector<int> num(n);
    for (int i = 0; i < n; i++)
    {
        cin >> num[i];
    }
 
    vector<int> dp(n, 1); 
 
    int ans = 1; 
 
    for (int i = 1; i < n; i++)
    {
        if (abs(num[i] - num[i-1]) <= 1)
        {
            dp[i] = dp[i - 1] + 1; 
            ans = max(ans, dp[i]); // 更新最大长度
        }
    }
 
    cout << ans;
 
    return 0;
}

B

还是那个排列函数

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#include<bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
const int N = 15;
int a[N];
string s;
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin>>s;
    int n = s.size();
    for(int i=0; i<n; i++)
        a[i] = s[i] - 'a';
    sort(a, a+n);
    int res = 0;
    do
    {
        bool flag = true;
        for(int i=1; i<n; i++)
        {
            if(a[i]==a[i-1])
            {
                flag = false;
                break;
            }
        }
        if(flag)
            res++;
    }while(next_permutation(a, a+n));
    cout<<res<<endl;
    return 0;
}

C

计算得到

\(z = y / (v + x*t) + t\)

该函数先单调递减，然后单调递增

三分！！

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#include<bits/stdc++.h>
#define int long long 
using namespace std;
const int N = 100010;
const double INF = 1e6+10;
double v0,x,y;
 
double time(double t)
{
    return y / (v0 + t * x) + t;
}
 
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    cin >> v0 >> x >> y;
    double l = 0,r = INF;
    for(int i=0;i<100;i++)
    {
        double mid = (r - l) / 3.0;
        double mid1 = l + mid;
        double mid2 = r - mid;
        if(time(mid1)>time(mid2)) l = mid1;
        else r = mid2;
    }
     
    printf("%.7lf",time(l));
    return 0;
}

求导O(1)也可以做

自己求（）

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#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long LL;
 
void solve()
{
    long double v0,x,y;
    cin >> v0 >> x >> y;
    if(v0 > sqrt(x*y)) printf("%.12llf",y/v0);
    else printf("%.12llf",2*sqrt(y/x) - v0/x);
}
 
int main() {
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int t = 1;
    while(t--) solve();
    return 0;
}

D

本题由于连续的一定是满足连续的情况的

有一个结论就是n的字符串一定是有n*(n+1)/2个回文串的

接下来只需要遍历左右就可以了

注意加上min和break

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#include <bits/stdc++.h>
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define frep(i, a, n) for (int i = a; i >= n; i--)
using ll = long long;
using namespace std;
#define int long long
const int N=1e6+100;
int mod=1e9+7;
int a[N];
signed main()
{
    int n;
    cin>>n;
    int ans=0;
    rep(i,1,n)
    {
        cin>>a[i];
        ans=(ans+(1+a[i])*a[i]/2)%mod;
    }
    rep(i,2,n)
    {
        int l=i-1;
        int r=i+1;
        while(l>=1&&r<=n)
        {
            ans=(ans+min(a[l],a[r]))%mod;
            if(a[l]!=a[r])
            {
                break;
            }
            l--;
            r++;
        }
    }
    cout<<ans;
    return 0;
}
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