题目链接

A. 牛牛的三角形

题意：

题解：


AC代码

/*
    Author:zzugzx
    Lang:C++
    Blog:blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};

int a[110];


int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            for(int k=1;k<=n;k++){
                if(i==k||i==j||k==j)continue;
                if(a[i]+a[j]>a[k]&&a[i]+a[k]>a[j]&&a[k]+a[j]>a[i]){
                    cout<<a[i]<<' '<<a[j]<<' '<<a[k]<<endl;
                    return 0;
                }
            }
    cout<<"No solution"<<endl;
    return 0;
}

B. 牛牛的鱼缸

题意：


题解：












AC代码

/*
    Author:zzugzx
    Lang:C++
    Blog:blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};



int main()
{
    //ios::sync_with_stdio(false);
    //cin.tie(0);cout.tie(0);
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    double h,l,H,L;
    cin>>h>>l>>H>>L;
    double x=H/L;
    if(x>=h/l)
        printf("%.9lf",h*h/x/2);
    else{
        double ans=(h-l*x+h)*l/2;
        printf("%.9lf",ans);
    }
    return 0;
}

C. 牛牛的揠苗助长

题意：





题解：













AC代码

/*
    Author:zzugzx
    Lang:C++
    Blog:blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};

int n;
ll a[maxn],b[maxn];
bool ok(ll x){
    for(int i=1;i<=n;i++){
        b[i]=a[i]+x/n;
        if(x%n>=i)b[i]++;
    }
    sort(b+1,b+1+n);
    ll y=b[(n+1)/2],res=0;
    for(int i=1;i<=n;i++)res+=abs(y-b[i]);
    return x>=res;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    ll l=0,r=1e18,ans;
    while(l<=r){
        ll mid=l+r>>1;
        if(ok(mid))ans=mid,r=mid-1;
        else l=mid+1;
    }
    cout<<ans;
    return 0;
}

D. 牛牛的01限定串

题意：








题解：








AC代码

/*
    Author:zzugzx
    Lang:C++
    Blog:blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};

ll dp1[1010][1010],dp2[1010][1010];
int sum[1010];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n,c0,c1,vp,vs;
    cin>>n>>c0>>c1>>vp>>vs;
    string s,t;
    cin>>s>>t;
    s='.'+s,t='.'+t;
    for(int i=1;i<=n;i++){
        sum[i]=sum[i-1];
        if(s[i]=='1')sum[i]++;
    }
    for(int i=0;i<=n;i++)
        for(int j=0;j<=c1;j++)
            dp1[i][j]=1e18,dp2[i][j]=-1e18;
    dp1[0][0]=dp2[0][0]=0;
    for(int i=1;i<=n;i++)
        for(int j=0;j<=c1;j++){
            ll tmp;
            if(t[i]!='1'){
                tmp=dp1[i-1][j];
                if(j==sum[i])tmp+=vp;
                if(sum[n]-sum[i-1]==c1-j)tmp+=vs;
                dp1[i][j]=min(dp1[i][j],tmp);
                tmp=dp2[i-1][j];
                if(j==sum[i])tmp+=vp;
                if(sum[n]-sum[i-1]==c1-j)tmp+=vs;
                dp2[i][j]=max(dp2[i][j],tmp);
            }
            if(t[i]!='0'&&j-1>=0){
                tmp=dp1[i-1][j-1];
                if(j==sum[i])tmp+=vp;
                if(sum[n]-sum[i-1]==c1-j+1)tmp+=vs;
                dp1[i][j]=min(dp1[i][j],tmp);
                tmp=dp2[i-1][j-1];
                if(j==sum[i])tmp+=vp;
                if(sum[n]-sum[i-1]==c1-j+1)tmp+=vs;
                dp2[i][j]=max(dp2[i][j],tmp);
            }
        }
    cout<<dp1[n][c1]<<' '<<dp2[n][c1];
    return 0;
}