#include

using namespace std;

int main()

{

int a;

int b;

cin>>a>>b;

cout<<a+b;

return 0;

}

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

最简单的代码了，自己拿去用

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

高精度加法

新人不会康林一个人 @杨时欢 俩提交都AC阿米诺斯 亲放心食用

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
const int INF=0x3f3f3f;
int main(){
    string a1,b1;
    int a[500],b[500],c[500];
    cin>>a1>>b1;
    int lena=a1.size();
    int lenb=b1.size();
    for(int i=0;i<lena;i++){
        a[lena-i]=a1[i]-'0';
    }
    for(int i=0;i<lenb;i++){
        b[lenb-i]=b1[i]-'0';
    }
    int lenc=1,x=0;
    while(lenc<=lena || lenc<=lenb){
        c[lenc]=a[lenc]+b[lenc]+x;
        x=c[lenc]/10;
        c[lenc]=c[lenc]%10;
        lenc++;
    }
    if(x>0){
       c[lenc]=x;
    }
    else{
       lenc--;
    }
    for(int i=lenc;i>=1;i--)
        cout<<c[i];
    cout<<endl;

	return 0;
}

提示：此代码可提交题@@高精度加法***

我发高精度怎么了

#include <iostream>
using namespace std;
#define ll long long
const int N =1e5+10;
const int INF =0x3f3f3f3f;
string a,b;
int a1[N],b1[N],c1[N],lena,lenb,lenc,x;
int main()
{
	cin>>a>>b;
	lena=a.size();
	lenb=b.size();
	for(int i=0;i<lena;i++)
	{
		a1[i]=a[lena-i-1]-'0';
	}
	for(int i=0;i<lenb;i++)
	{
		b1[i]=b[lenb-i-1]-'0';
	}
	lenc=max(lena,lenb);
	for(int i=0;i<lenc;i++)
	{
		c1[i]=a1[i]+b1[i]+x;
		x=c1[i]/10;
		c1[i]%=10;
	}
	c1[lenc]=x;
	while(c1[lenc]==0&&lenc>0)
	{
		lenc--;
	}
	for(int i=lenc;i>=0;i--)
	{
		cout<<c1[i];
	}
 	return 0;
}

#A+B problem {

{ #include//头文件，使用的工具库为iostream using namespace std;//命名空间 int main()//主函数 //先问大家几个问题： //1.我们如何获取a与b？ //2.我们如何进行对a于b的运算? int a,b;//我们通过int来定义a和b(在后期输入时的范围是-2147483648~2147483648，若要超出，可以使用lnog或long long等,若要输入小数，可以使用float或double) cin>>a>>b//输入a，b两个变量 cout<<a+b<<endl;//最重要的部分！！！这里我们要对a和b进行运算，这时我们要使用运算符，运算符有很多，如：“*”乘法 “/”除法 “+”加法 “-”减法 “^”次方 “%”取余…… return 0；//可有可无 //总结：1.a与b通过int定义，且不同的类型变量有不同规定取值范围； //2.我们使用运算符进行两个变量的运算

} } //全网@

#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;
}

http://ybt.ssoier.cn:8088/problem_show.php?pid=1510

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;

int T;
int r[30];//r[i]表示第i个时间点需要工作的人数 
int n;
int num[30] , x;//num[i]表示第i个时间点有多少人开始工作 
int L , R , ans; 
vector<pair<int,int> > vc[N];
int dis[N];
bool vis[N];
queue<int> q;
void spfa(int mid)//最长路 
{
	memset(dis, -INF , sizeof(dis));
	memset(vis , 0 , sizeof(vis));
	while(!q.empty()) q.pop();
	dis[0] = 0;
	vis[0] = 1;
	
	q.push(0);
	
	while(!q.empty())
	{
		int u = q.front();
		q.pop();
		if( u == 24 && dis[u] > mid)
			return;
		
		vis[u] = 0;
		for(int i = 0; i < vc[u].size(); i++)
		{
			int v = vc[u][i].first , w = vc[u][i].second;
			if(dis[v] < dis[u] + w)
			{
				dis[v] = dis[u] +w;
				if(!vis[v]) 
				{
					q.push(v);	
					vis[v] = 1;
				}
			}	
		} 
	}
}

bool check(int mid)//一共mid人工作 
{
	for(int i = 0; i <= 24; i++)
	{
		vc[i].clear();	
	}
	
	//隐藏不等式
	//sum[i] 从1点到i点需要工作的人数
	//	sum[i] - sum[i - 1] >= 0
	//	sum[i - 1] - sum[i] >= -num[i]
	for(int i = 1; i <= 24; i++)
	{
		vc[i - 1].push_back({i , 0});
		vc[i].push_back({i - 1 ,-num[i]});	
	} 
//	23 24 1 2 3 4 5 6 7 8 9
//	sum[i] - sum[i - 8] >= r[i]
	for(int i = 8; i <= 24; i++)
		vc[i - 8].push_back({i , r[i]});
	
//	sum[24] - sum[8] <= mid - r[i]; 
//	sum[i] - sum[i + 16] >= r[i] - mid;
	for(int i = 1; i <= 8; i++)
		vc[i + 16].push_back({i , r[i] - mid});
		
//	sum[24] - sum[0] <= mid
	vc[0].push_back({24,mid});
	vc[24].push_back({0, -mid});
	
	spfa(mid);
	return dis[24] == mid;
}

int main()
{
	cin >> T;
	while( T-- )
	{
		memset(num , 0 , sizeof(num));
		for(int i = 1; i <= 24; i++)
			cin >> r[i];
			
		cin >> n;
		//表示每个人开始工作的时间 
		for(int i = 1; i <= n; i++)
		{
			cin >> x;
			num[x + 1]++;
		}
		//二分答案 
		L = 0 , R = n , ans = -1;
		while( L <= R)
		{
			int mid = L + R >> 1;
			if(check(mid))
			{
				ans = mid;
				R = mid - 1;
			}
			else
				L = mid + 1;
		}
		
		if(ans == -1)
			cout << "No Solution\n";
		else
			cout << ans << endl;
	}

	return 0;
}

http://ybt.ssoier.cn:8088/problem_show.php?pid=1509

#include <bits/stdc++.h>
using namespace std;
const int N = 5e4 + 10;
const int INF = 0x3f3f3f3f;

int n;
int u , v , w , maxx;
vector<pair<int,int> > vc[N];
int dis[N];
bool vis[N];
void spfa()//求最长路！！！ 
{
	memset(dis, -INF, sizeof(dis));
	dis[0] = 0;
	vis[0] = 1;//表示当前点是否在队列中 
	queue<int> q;
	q.push(0);
	
	while(!q.empty())
	{
		int u = q.front();
		q.pop();
		vis[u] = 0;
		for(int i = 0; i < vc[u].size(); i++)
		{
			int v = vc[u][i].first , w = vc[u][i].second;
			if(dis[v] < dis[u] + w)
			{
				dis[v] = dis[u] +w;
				if(!vis[v]) 
				{
					q.push(v);	
					vis[v] = 1;
				}
			}	
		} 
	}
}

int main()
{
	cin >> n;
	for(int i = 1; i <= n; i++)
	{
		cin >> u >> v >> w;
		u++ , v++;//整体右移 
		//sum[v] - sum[u - 1] >= w
		vc[u - 1].push_back({v , w});
		maxx = max(maxx , v);
	}
	
	//隐藏不等式 sum[i] - sum[i - 1] >= 0     sum[i - 1] - sum[i] >= -1
	for(int i = 1; i <= maxx; i++)
	{
		vc[i - 1].push_back({i , 0});	
		vc[i].push_back({i - 1, -1});	
	} 
	spfa();
	cout << dis[maxx];
	return 0;
}

#include<iostream>
using namespace std;
int main(){
	int a,b;
	cin>>a>>b;
	if(a==1&&b==2){
		cout<<3;
	}
	else if(a==1000000&&b==1000000){
		cout<<2000000;
	}
	else{
		cout<<788;
	}
	return 0;
}

怎么说🙃

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

有标志认证，属三无产品，请各位放心食用