https://ac.nowcoder.com/acm/contest/93268
A

# include <bits/stdc++.h>
using namespace std;
#define int long long 
signed main()
{
int T;
    cin>>T;
    while(T--)
    {
    string s;
    cin>>s;
    cout<<"Welcome to GDUT ";
    cout<<s<<endl;
    }
}

B
贪心即可

# include <bits/stdc++.h>
using namespace std;
#define int long long 
const int N=1e3+10;
int a[N],n,x;
signed main()
{
    cin>>n>>x;
    for(int i=1;i<=n;i++)
    cin>>a[i];
    sort(a+1,a+n+1);
    int k=x;
    int ans=0;
    for(int i=n;i>=1;i--)
    {
            if(x<=0)break;
    if(x>0)
    {
    x=x-a[i];
    ans++;
    }
    }
    cout<<ans<<" ";
    if(k<=n)
    {
           cout<<k;
           return 0;
    }
    else cout<<n;
}

C bfs

# include <bits/stdc++.h>
using namespace std;
#define int long long 
const int N=1e3+10;
int a[N],n,x;
signed main()
{
    cin>>n>>x;
    for(int i=1;i<=n;i++)
    cin>>a[i];
    sort(a+1,a+n+1);
    int k=x;
    int ans=0;
    for(int i=n;i>=1;i--)
    {
            if(x<=0)break;
    if(x>0)
    {
    x=x-a[i];
    ans++;
    }
    }
    cout<<ans<<" ";
    if(k<=n)
    {
           cout<<k;
           return 0;
    }
    else cout<<n;


}

E
二分答案 注意bool函数中一旦a[i]>mid就返回

# include <bits/stdc++.h>
using namespace std;
#define int long long 
int n,k;
const int N=1e6+10;
int a[N];
bool check(int mid)
{
    int ans=0,sum=0;
    for(int i=1;i<=n;i++)
    {
    if(a[i]>mid)return false;
    if(sum+a[i]>mid)
    {
    sum=0;
    ans++;
    }
    sum+=a[i];
    }
    if(sum)
    ans++;
    if(ans<=k)
    return true;
    return false;
}
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    cin>>n>>k;
    for(int i=1;i<=n;i++)
        cin>>a[i];
        int l=1,r=1e18;
        while(l<r)
        {
        int mid=l+r>>1;
        if(check(mid))r=mid;
        else l=mid+1;
        }
        cout<<l<<endl;
    }
}

F
简单dp
f[i]代表血量为i时所需的最小气力 可从三个方向转移
注意血量可能从负数转移过来

# include <bits/stdc++.h>
using namespace std;
#define int long long 
int hp,a,b,c,d;
const int N=1e6+01;
int k[N],f[N];
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    cin>>hp>>a>>b>>c>>d;
    for(int i=1;i<=hp;i++)
    cin>>k[i];
    //f[0]=1;
    for(int i=1;i<=hp;i++)
        f[i]=1e18;
    f[0]=0;
    for(int i=1;i<=hp;i++)
    {
    f[i]=min(f[i],f[max(i-b,0ll)]+a);
    f[i]=min(f[i],f[max(i-d,0ll)]+c);
    if(i-4*b>=0)
    f[i]=min(f[i],f[max(0ll,i-4*b-k[i-4*b]-d)]+4*a+c);
    }
    cout<<f[hp];
    cout<<endl;
    }
}

I
建图之后， 该更新k点，如果k此时不是最小值的话，那么无需看是否需要传送 ，是最小值的话（有多条最小值路线，就找传送次数最少得那条路线）多维护一个cnt即可

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
#define fi first
#define se second
#define i128 __int128
const int N = 2e5 + 10;
typedef pair<int, int> PII;
const int mod = 998244353;
const int inf = 1e13;
const int hash_mod = 1e9 + 7;
const double pi = 3.1415926535;
const double eps = 1e-9;
struct node{
    int x,dis;
    bool operator<(const node&b)const{
        return b.dis<dis;
    }
};
void solve() {
    int n,m,k,w;cin>>n>>m>>k>>w;
    vector<array<int,3>>g[n+1];
    for(int i=1;i<=m;i++){
        int x,y,z;cin>>x>>y>>z;
        g[x].push_back({y,z,0});
        g[y].push_back({x,z,0});
    }
    for(int i=1;i+k<=n;i++){
        g[i].push_back({i+k,w,1});
        g[i+k].push_back({i,w,1});
    }
    vector<int>dp(n+1,inf);
    vector<int>cnt(n+1,0);
    dp[1]=0;
    priority_queue<node>q;
    q.push({1,0});
    vector<int>vis(n+1,0);
    while(!q.empty()){
        auto [x,len]=q.top();
        q.pop();
        if(vis[x])continue;
        vis[x]=1;
        for(auto [v,w,st]:g[x])
        {
            if(dp[v]>dp[x]+w&&!vis[v])
            {
                dp[v]=dp[x]+w;
                cnt[v]=cnt[x]+st;
                q.push({v,dp[v]});
            }
            else if(dp[v]==dp[x]+w)
            {
                if(cnt[v]>cnt[x]+st)
                {
                    cnt[v]=cnt[x]+st;
                }
            }
        }       
    }
    if(dp[n]==inf)cout<<"-1 -1\n";
    else cout<<dp[n]<<" "<<cnt[n]<<endl;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int _ = 1;
    cin >> _;
    cout << setprecision(12);
    while (_--)solve();
    return 0 ^ 0;
}