C - Separated Lunch

题目：

思路：

dfs枚举每个数是否选入a数组中，求和比较

代码：

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N=25;

int a[N];
bool st[N];
int mn=0x3f3f3f3f;
int sum;
int n;
void dfs(int u, int s)
{
    if(u==n){
        mn=min(mn,max(sum-s,s));
        return;
    }
    dfs(u+1,s);
    dfs(u+1,s+a[u]);
}

int main()
{
    cin>>n;

    for(int i=0; i<n; i++){
        cin>>a[i];
        sum+=a[i];
    }
    dfs(0,0);
    cout<<mn<<'\n';

    return 0;

}

欣赏一下jiangly的代码：

#include <bits/stdc++.h>

using i64 = long long;
using u64 = unsigned long long;
using u32 = unsigned;

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    
    int N;
    std::cin >> N;
    
    std::vector<int> K(N);
    for (int i = 0; i < N; i++) {
        std::cin >> K[i];
    }
    
    int tot = std::accumulate(K.begin(), K.end(), 0);
    
    int ans = tot;
    std::vector<int> sum(1 << N);
    for (int s = 0; s < (1 << N); s++) {
        if (s > 0) {
            int u = __builtin_ctz(s);
            sum[s] = K[u] + sum[s ^ (1 << u)];
        }
        ans = std::min(ans, std::max(sum[s], tot - sum[s]));
    }
    std::cout << ans << "\n";
    
    return 0;
}

D - Laser Marking 

题目：

思路：

dfs一下1~n的全排列，注意每次递归两个端点，取最小值

代码：

#include <bits/stdc++.h>
#define fi first
#define se second;

using namespace std;

typedef long long LL;
typedef pair<int,int> PII;

int main()
{
    int n;
    cin>>n;

    int s, t;
    cin>>s>>t;
    vector<int> A(n), B(n), C(n), D(n);

    for(int i=0; i<n; i++){
        cin>>A[i]>>B[i]>>C[i]>>D[i];
    }

    double ans=1e18;
    vector<bool> vis(n);
    auto dfs = [&] (auto &self ,int x,int y, double sum=0.0, int c=0){
        if(c==n){
            ans= min(ans,sum);
            return;
        }
        for(int i=0; i<n; i++){
            if(vis[i]){
                continue;
            }
            double d1=hypot(A[i]-x, B[i]-y);
            double d2=hypot(C[i]-x, D[i]-y);
            double d0=hypot(A[i]-C[i], B[i]-D[i]);
            vis[i]=true;
            self(self ,C[i], D[i], sum+d1/s +d0/t, c+1);
            self(self ,A[i], B[i], sum+d2/s +d0/t, c+1);
            vis[i]=false;
        }
    };
    dfs(dfs,0,0);
    cout<<fixed<<setprecision(20)<<ans<<'\n';

    return 0;
}

E - Sensor Optimization Dilemma 2

题目：

思路：

截取了两位博主的思路

代码：

#include <bits/stdc++.h>
#define fi first
#define se second;

using namespace std;

typedef long long LL;
typedef pair<int,int> PII;

int main()
{
    int n,x;
    cin>>n>>x;
    vector<int> A(n), B(n), P(n), Q(n);
    for(int i=0; i<n; i++){
        cin>>A[i]>>P[i]>>B[i]>>Q[i];
        if(A[i]*Q[i]<B[i]*P[i]){
            swap(A[i], B[i]);
            swap(P[i], Q[i]);
        }
    }

    auto check = [&](int u){
        LL ans=0;
        for(int i=0; i<n; i++){
            LL res=1e18;
            for(int j=0; j<=100; j++){
                LL v=j*Q[i];
                int need =u-j*B[i];
                if(need > 0){
                    v+=1LL *(need + A[i]-1)/A[i] *P[i];
                }
                res=min(res,v);
            }
            ans+=res;
        }
        return ans <= x;
    };
    int l=0, r=1e9;
    while(l<r){
        int mid=(l+r+1)/2;
        if(check(mid)){
            l=mid;
        } else {
            r=mid-1;
        }
    }
    cout<<l<<'\n';

    return 0;
}