64 条题解

  • -2
    @ 2026-5-15 17:25:00

    #include <bits/stdc++.h> using namespace std; int main(){ int a; int b; cin>>a>>b; cout<<a+b; return 0; }

    • -2
      @ 2026-4-15 13:23:53

      python代码

      
      a=int(input())
      b=int(input())
      print(a+b)
      
      • -2
        @ 2026-3-29 19:04:19

        #include <bits/stdc++.h> using namespace std; int main(){ int a,b,s; cin>>a>>b; s=a+b; cout<<s; return 0; }

        • -2
          @ 2026-3-27 15:15:43
          #include <bits/stdc++.h>
          using namespace std;
          
          int main(){
              int a, b;
              cin >> a >> b;
              cout << a + b << endl;
              return 0;
          }
          
          • -2
            @ 2026-3-22 17:42:07
            #include <bits/stdc++.h>
            using namespace std;
            int main(){
              int a,b,s;
              cin>>a>>b;
              s=a+b;
              cout<<s;
              return 0;
            }
            
            • -2
              @ 2026-3-16 21:13:55

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

              int main(){ int a,b; cin>>a>>b; cout<<a+b; return 0; }

              • -2
                @ 2026-2-2 12:29:27
                #include<bits/stdc++.h>
                using namespace std;
                
                const int N=1e5+5;
                
                int n , m , a[N];
                int u , v;
                vector<int> vc[N];
                int dfn[N] , low[N] , cnt;
                bool vis[N];
                stack<int> st;
                int belong_cnt; //强连通分量个数 
                vector<int> belong[N];//强连通分量 
                int color[N];//color[i]表示第i头牛所处的强连通分量  
                vector<int> newvc[N];
                int in[N];//入度 
                int ans[N];
                void tarjan(int u)
                {
                	dfn[u] = low[u] = ++cnt;
                	vis[u] = 1;
                	st.push(u);
                	for(int i = 0; i < vc[u].size(); i++)
                	{
                		int v = vc[u][i];
                		if(!dfn[v])
                		{
                			tarjan(v);
                			low[u] = min(low[u] , low[v]);
                		}
                		else if(vis[v])
                		{
                			low[u] = min(low[u] , low[v]);
                		}	
                	}
                	//强连通分量到头	
                	if(dfn[u] == low[u])
                	{
                		while(!st.empty())
                		{
                			v = st.top();
                			st.pop();
                			vis[v] = 0;
                			color[v] = u;//当前牛所处的强连通分量 
                			if(u == v) break;
                			a[u] += a[v];
                		}
                	}
                }
                
                void tupo()
                {
                	queue<int> q;
                	for(int i = 1; i <= n; i++)
                	{
                		if(color[i] == i && !in[i])
                		{
                			q.push(i);
                			ans[i] = a[i];
                		}
                	}
                	
                	while(!q.empty())
                	{
                		int top = q.front();
                		q.pop();
                		
                		for(int i = 0; i < newvc[top].size(); i++)
                		{
                			v = newvc[top][i];
                			ans[v] = max(ans[v] , ans[top] + a[v]);
                			in[v]--;
                			if(!in[v])
                				q.push(v);
                		}
                	}
                	
                	int maxx = 0;
                	for(int i = 1; i <= n; i++)
                		maxx = max(maxx , ans[i]);
                	cout << maxx;
                }
                
                int main(){
                	cin >> n >> m;
                	for(int i = 1; i <= n; i++)
                		cin >> a[i];
                	while( m-- )
                	{
                		cin >> u >> v;
                		vc[u].push_back(v); 
                	}
                	for(int i = 1; i <= n; i++)
                		if(!dfn[i])
                			tarjan(i);
                	
                	//重新建边
                	for(int i = 1; i <= n; i++)
                	{
                		for(int j = 0; j < vc[i].size(); j++)
                		{
                			v = vc[i][j];
                			if(color[i] != color[v])
                			{
                				newvc[color[i]].push_back(color[v]);
                				in[color[v]]++; 
                			}	
                		}	
                	} 
                	
                	tupo();
                	
                	return 0;
                
                • -2
                  @ 2026-2-2 12:28:28
                  #include<bits/stdc++.h>
                  using namespace std;
                  
                  const int N=1e5+5;
                  
                  int n , m , a[N];
                  int u , v;
                  vector<int> vc[N];
                  int dfn[N] , low[N] , cnt;
                  bool vis[N];
                  stack<int> st;
                  int belong_cnt; //强连通分量个数 
                  vector<int> belong[N];//强连通分量 
                  int color[N];//color[i]表示第i头牛所处的强连通分量  
                  vector<int> newvc[N];
                  int in[N];//入度 
                  int ans[N];
                  void tarjan(int u)
                  {
                  	dfn[u] = low[u] = ++cnt;
                  	vis[u] = 1;
                  	st.push(u);
                  	for(int i = 0; i < vc[u].size(); i++)
                  	{
                  		int v = vc[u][i];
                  		if(!dfn[v])
                  		{
                  			tarjan(v);
                  			low[u] = min(low[u] , low[v]);
                  		}
                  		else if(vis[v])
                  		{
                  			low[u] = min(low[u] , low[v]);
                  		}	
                  	}
                  	//强连通分量到头	
                  	if(dfn[u] == low[u])
                  	{
                  		while(!st.empty())
                  		{
                  			v = st.top();
                  			st.pop();
                  			vis[v] = 0;
                  			color[v] = u;//当前牛所处的强连通分量 
                  			if(u == v) break;
                  			a[u] += a[v];
                  		}
                  	}
                  }
                  
                  void tupo()
                  {
                  	queue<int> q;
                  	for(int i = 1; i <= n; i++)
                  	{
                  		if(color[i] == i && !in[i])
                  		{
                  			q.push(i);
                  			ans[i] = a[i];
                  		}
                  	}
                  	
                  	while(!q.empty())
                  	{
                  		int top = q.front();
                  		q.pop();
                  		
                  		for(int i = 0; i < newvc[top].size(); i++)
                  		{
                  			v = newvc[top][i];
                  			ans[v] = max(ans[v] , ans[top] + a[v]);
                  			in[v]--;
                  			if(!in[v])
                  				q.push(v);
                  		}
                  	}
                  	
                  	int maxx = 0;
                  	for(int i = 1; i <= n; i++)
                  		maxx = max(maxx , ans[i]);
                  	cout << maxx;
                  }
                  
                  int main(){
                  	cin >> n >> m;
                  	for(int i = 1; i <= n; i++)
                  		cin >> a[i];
                  	while( m-- )
                  	{
                  		cin >> u >> v;
                  		vc[u].push_back(v); 
                  	}
                  	for(int i = 1; i <= n; i++)
                  		if(!dfn[i])
                  			tarjan(i);
                  	
                  	//重新建边
                  	for(int i = 1; i <= n; i++)
                  	{
                  		for(int j = 0; j < vc[i].size(); j++)
                  		{
                  			v = vc[i][j];
                  			if(color[i] != color[v])
                  			{
                  				newvc[color[i]].push_back(color[v]);
                  				in[color[v]]++; 
                  			}	
                  		}	
                  	} 
                  	
                  	tupo();
                  	
                  	return 0;
                  
                  • @ 2026-5-31 17:30:55

                    老师,最后的括号没了 应该在“return 0”下边的

                • -2
                  @ 2025-5-24 15:28:36
                  #include<bits/stdc++.h>//万能头文件
                  using namespace std;
                  
                  int main(){
                  //定义int类型变量a,b
                  int a;
                  int b;
                  //输入变量a,b
                  scanf("%d",&a);
                  scanf("%d",&b);
                  //输出a,b
                  printf(" %d\n", a + b);
                  //exit(0); 或 return 0; 结束程序
                  return 0;
                  }
                  
                  
                  • -2
                    @ 2025-5-11 9:37:50

                    权威

                    #include<iostream>
                    #include<cstring>
                    #include<cstdio>
                    #include<cstring>
                    using namespace std;
                    struct node 
                    {
                        int data,rev,sum;
                        node *son[2],*pre;
                        bool judge();
                        bool isroot();
                        void pushdown();
                        void update();
                        void setson(node *child,int lr);
                    }lct[233];
                    int top,a,b;
                    node *getnew(int x)
                    {
                        node *now=lct+ ++top;
                        now->data=x;
                        now->pre=now->son[1]=now->son[0]=lct;
                        now->sum=0;
                        now->rev=0;
                        return now;
                    }
                    bool node::judge(){return pre->son[1]==this;}
                    bool node::isroot()
                    {
                        if(pre==lct)return true;
                        return !(pre->son[1]==this||pre->son[0]==this);
                    }
                    void node::pushdown()
                    {
                        if(this==lct||!rev)return;
                        swap(son[0],son[1]);
                        son[0]->rev^=1;
                        son[1]->rev^=1;
                        rev=0;
                    }
                    void node::update(){sum=son[1]->sum+son[0]->sum+data;}
                    void node::setson(node *child,int lr)
                    {
                        this->pushdown();
                        child->pre=this;
                        son[lr]=child;
                        this->update();
                    }
                    void rotate(node *now)
                    {
                        node *father=now->pre,*grandfa=father->pre;
                        if(!father->isroot()) grandfa->pushdown();
                        father->pushdown();now->pushdown();
                        int lr=now->judge();
                        father->setson(now->son[lr^1],lr);
                        if(father->isroot()) now->pre=grandfa;
                        else grandfa->setson(now,father->judge());
                        now->setson(father,lr^1);
                        father->update();now->update();
                        if(grandfa!=lct) grandfa->update();
                    }
                    void splay(node *now)
                    {
                        if(now->isroot())return;
                        for(;!now->isroot();rotate(now))
                        if(!now->pre->isroot())
                        now->judge()==now->pre->judge()?rotate(now->pre):rotate(now);
                    }
                    node *access(node *now)
                    {
                        node *last=lct;
                        for(;now!=lct;last=now,now=now->pre)
                        {
                            splay(now);
                            now->setson(last,1);
                        }
                        return last;
                    }
                    void changeroot(node *now)
                    {
                        access(now)->rev^=1;
                        splay(now);
                    }
                    void connect(node *x,node *y)
                    {
                        changeroot(x);
                        x->pre=y;
                        access(x);
                    }
                    void cut(node *x,node *y)
                    {
                        changeroot(x);
                        access(y);
                        splay(x);
                        x->pushdown();
                        x->son[1]=y->pre=lct;
                        x->update();
                    }
                    int query(node *x,node *y)
                    {
                        changeroot(x);
                        node *now=access(y);
                        return now->sum;
                    }
                    int main()
                    {
                        scanf("%d%d",&a,&b);
                        node *A=getnew(a);
                        node *B=getnew(b);
                        //连边 Link
                            connect(A,B);
                        //断边 Cut
                            cut(A,B);
                        //再连边orz Link again
                            connect(A,B);
                        printf("%d\n",query(A,B)); 
                        return 0;
                    }
                    
                    
                    
                    
                    • -3
                      @ 2026-5-31 17:29:11

                      AC干活

                      #include <bits/stdc++.h>
                      int main(){
                      	int a,b;
                      	std::cin>>a>>b;
                      	std::cout<<a+b; 
                      }
                      

                      拿走不用谢(●'◡'●)

                      • -3
                        @ 2026-4-27 21:14:43

                        #include using namespace std; int main(){ int a,b; cin>>a>>b; cout<<a+b; return 0; }

                        • -3
                          @ 2026-4-25 18:01:58

                          #include<bits/stdc++.h> using namespace std; const int N=1010;//1表示开头为1,2表示10的2次方 const int INT=0x3f3f3f3f;//INT+INT int范围内最大INT*INT ,long long; int n,m; void dfs(int n,int m){ int sum=0,ans=0; sum=n; ans=m; cout<<ans+1-1+1-1+1-1+sum+1-1+1-1+1-1; } int main( ) { cin>>n>>m; dfs(n,m); }

                          • -3
                            @ 2026-4-8 19:18:42
                            
                            ```#include <bits/stdc++.h>
                            using namespace std;
                            int main(){
                              int a,b,s;
                              cin>>a>>b;
                              s=a+b;
                              cout<<s;
                              return 0;
                            }
                            • -3
                              @ 2026-4-7 13:02:50

                              #include using namespace std; int main() { int a,b; cin>>a>>b; cout<<a+b; return 0; }

                              • -3
                                @ 2025-10-25 9:34:12
                                #include<iostream>
                                using namespace std;
                                int main(){
                                    int a,b;
                                    cin>>a>>b;
                                    cout<<a+b;
                                return 0;
                                }
                                
                                • -3
                                  @ 2025-2-21 19:53:22

                                  最短题解

                                  #include<iostream>
                                  int a,b;int main(){std::cin>>a>>b;std::cout<<a+b;}
                                  
                                  • -3
                                    @ 2025-1-23 11:13:08
                                    #include<iostream>
                                    using namespace std;
                                    int main()
                                    {
                                    	int a,b;
                                    	cin>>a>>b;
                                    	cout<<a+b;
                                    }
                                    
                                    • -3
                                      @ 2024-11-16 16:21:16
                                      #include<iostream>
                                      using namespace std;
                                      int main(){
                                      	int a,b,c;
                                      	cin>>a>>b;
                                      	c=a+b;
                                      	cout<<c;
                                      }
                                      
                                      • -4
                                        @ 2025-12-30 22:23:26
                                        #include<iostream>
                                        using namespace std;
                                        int a,b;
                                        int main ( ) {
                                            cin>>a>>b;
                                            cout<<a+b;
                                        }
                                        

                                        信息

                                        ID
                                        1
                                        时间
                                        1000ms
                                        内存
                                        128MiB
                                        难度
                                        1
                                        标签
                                        递交数
                                        5264
                                        已通过
                                        1486
                                        上传者