46 条题解

  • 1
    @ 2026-4-15 13:26:15 
    #include<stdio.h>
int main(){
    int a,b;
    scanf("%d%d",&a,&b);
    printf("%d",a+b);
}
    • 1
      @ 2026-4-15 13:23:53 
      
a=int(input())
b=int(input())
print(a+b)
      • 1
        @ 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;
}
        • 1
          @ 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;
}
          • 1
            @ 2025-10-25 9:33:33 
            #include<bits/stdc++.h>
using namespace std;
const int N=1e3+10;
int a,b;
int main()
{
	cin >> a >> b;
	cout << a+b;
	return 0;
}
            • 0
              @ 2026-4-7 13:02:50

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

              • 0
                @ 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;
}
                • 0
                  @ 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;
                  • 0
                    @ 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;
                    • 0
                      @ 2025-12-14 12:27:47 
                      #include <iostream>
using namespace std;
int main()
{
	int a,b;
	cin>>a>>b;
	cout<<a+b;
	return 0;
}
                      • 0
                        @ 2025-11-22 9:37:06 
                        #include<bits/stdc++.h>
using namespace std;
const int N=1e3+10;
int Accepted(int a,int b){//Accepted()
	return a+b; 
}
int main()
{
	int a,b;
	cin >> a >> b;
	cout << Accepted(a,b);
	return 0;
}
//老登布置的作业系列
                        • 0
                          @ 2025-11-8 9:43:09 
                          #include<bits/stdc++.h>

using namespace std;
int main()
{
	string a1,b1;
	int a[10000]={},b[10000]={},c[10000]={};
	cin >> a1 >> b1;
	int lena=a1.size();
	int lenb=b1.size();
	for(int i=1;i<=lena;i++)
	{
		a[i]=a1[lena-i]-'0';
	} 
	for(int i=1;i<=lenb;i++)
	{
		b[i]=b1[lenb-i]-'0';
	}  
	int lenc=1;
	while(lenc<=lena||lenc<=lenb)
	{
		c[lenc]+=a[lenc]+b[lenc];
		if(c[lenc]>9)
		{
			c[lenc]-=10;
			c[lenc+1]++; 
		}
		lenc++;
	} 
	if(c[lenc]==0)
	{
		lenc--;
	}
	for(int i=lenc;i>=1;i--)
	{
		cout << c[i];
	}
	return 0;
}
                          • -1
                            @ 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; }

                            • -1
                              @ 2024-7-26 10:58:29

                              A+B Problem题解

                              新用户强烈建议阅读此帖

                              首先我们要理清思路

                              1.需要用到什么样的头文件？

                              2.用什么样的数据范围？

                              3.思路是什么？

                              首先题目中的数据范围是1≤a,b≤10^6, 而int 的范围是-2147483648-2147483647 正合题意，所以数据类型可以用int

                              话不多说，直接上代码

                              #include<iostream>//导入头文件，iostream里面是标准输入输出流（我说的什么？） 
using namespace std;//使用标准命名空间 
int main(){//主函数，程序的入口 
	int a,b;//创建a,b两个整型变量 
	cin>>a>>b;//输入 a , b 两个变量 
	cout<<a+b; //输出a+b的内容 
	return 0; 
}

                              本蒟蒻发的第一篇题解，请多多支持喵~~

                              • -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
                                  @ 2025-12-30 22:23:26 
                                  #include<iostream>
using namespace std;
int a,b;
int main ( ) {
    cin>>a>>b;
    cout<<a+b;
}
                                  • -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
                                        @ 2026-1-31 8:59:47

                                        1行。

                                        int main() { int a,b; __builtin_scanf("%d%d",&a,&b); __builtin_prinf("%d",a+b);
                                        • -3
                                          @ 2025-12-28 18:10:49

                                          #include<bits/stdc++.h> using namespace std; const int N=1e3+10; int a,b; int main() { cin >> a >> b; cout << a+b; return 0; }

                                          A + B Problem

                                          信息

                                          ID
                                          1
                                          时间
                                          1000ms
                                          内存
                                          128MiB
                                          难度
                                          1
                                          标签
                                          递交数
                                          5148
                                          已通过
                                          1471
                                          上传者
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