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

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

额...... 这道题没什么可讲的吧......

#include <bits/stdc++.h>
using namespace std;
int main()
{
    long long a,b;//十年OI一场空，不开long long见祖宗
    cin>>a>>b;
    cout<<a+b<<endl;
    return 0;
}

简单题

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

世界上最好的题解😄

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

看完的麻烦点个赞呗👀️

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

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

直接到洛谷上找:

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

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
using namespace std;
const int N = 1005;
struct bign
{
    int len,s[N];
    bign()  {  memset(s,0,sizeof(s));  len=1;  }
    bign(int num)  {  *this=num; }
    bign(char *num) { *this=num; }
    bign operator =(int num)
    {
        char c[N];
        sprintf(c,"%d",num);
        *this=c;
        return *this;
    }
    bign operator =(const char *num)
    {
        len=strlen(num);
        for (int i=0;i<len;i++) s[i]=num[len-1-i]-'0';
        return *this;
    }
    string str()
    {
        string res="";
        for (int i=0;i<len;i++) res=(char)(s[i]+'0')+res;
        return res;
    }
    void clean()
    {
        while (len>1&&!s[len-1]) len--;
    }
    bign operator +(const bign &b)
    {
        bign c;    
        c.len=0;
        for (int i=0,g=0;g||i<len||i<b.len;i++)
        {
            int x=g;
            if (i<len) x+=s[i];
            if (i<b.len) x+=b.s[i];
            c.s[c.len++]=x%10;
            g=x/10;
        }
        return c;
    }
    bign operator -(const bign &b)
    {
        bign c;
        c.len=0;
        int x;     
        for (int i=0,g=0;i<len;i++)
        {
            x=s[i]-g;
            if (i<b.len) x-=b.s[i];
            if (x>=0) g=0;
            else{          
                x+=10;
                g=1;
            };
            c.s[c.len++]=x;
        }
        c.clean();
        return c;
    }
    bign operator *(const bign &b)
    {
        bign c;
        c.len=len+b.len;
        for (int i=0;i<len;i++) for (int j=0;j<b.len;j++) c.s[i+j]+=s[i]*b.s[j];
        for (int i=0;i<c.len-1;i++) { c.s[i+1]+=c.s[i]/10; c.s[i]%=10; }
        c.clean();
        return c;  
    }
    bool operator <(const bign &b)
    {
        if (len!=b.len) return len<b.len;
        for (int i=len-1;i>=0;i--)
             if (s[i]!=b.s[i]) return s[i]<b.s[i];
        return false;
    }
    bign operator +=(const bign &b)
    {
        *this=*this+b;
        return *this;
    }
    bign operator -=(const bign &b)
    {
        *this=*this-b;
        return *this;
    }  
};
istream& operator >>(istream &in,bign &x)
{
  string s;
  in>>s;
  x=s.c_str();
  return in;
}
ostream& operator <<(ostream &out,bign &x)
{
    out<<x.str();
    return out;
}
int main(){
    bign a,b,c;
    ios::sync_with_stdio(false);
    cin>>a>>b;
//    cout<<a<<endl;
//    cout<<b<<endl;
    c=a+b;
    cout<<c<<endl;
    return 0;
}

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
struct node{
    int str[26];
    int sum;
}s[1000];
char str1[100];
int t=0,tot=0,ss=0;
bool f1;
void built()
{
    t=0;
    for(int i=0;i<strlen(str1);i++)
    {
         if(str1[i]=='-'){
             f1=true;continue;
         }
         if(!s[t].str[str1[i]-'0'])
         s[t].str[str1[i]-'0']=++tot;
         t=s[t].str[str1[i]-'0'];
         s[t].sum=str1[i]-'0';
    }
}
int query()
{
   int t=0;int s1=0;
   for(int i=0;i<strlen(str1);i++)
   {
           if(str1[i]=='-') continue;
           if(!s[t].str[str1[i]-'0']) return s1;
           t=s[t].str[str1[i]-'0'];
           s1=s1*10+s[t].sum;
   }
   return s1;
}
int main()
{    
  for(int i=1;i<=2;i++)
  {
      f1=false;
      scanf("%s",str1);
    built();
    if(f1)
      ss-=query();
      else ss+=query();
  }
  printf("%d",ss);
  return 0;    
}

Description

输入两个数字 $a,b$，输出 $a + b$。

Solution

对 $\text{OI}$ 新手的提醒：

• 记得开 unsigned long long！
• 主函数一定要 return 0！

Code

#include <bits/stdc++.h>

using namespace std;

int main() {
   unsigned long long a, b;
   cin >> a >> b;
   cout << a + b << '\n';
   return 0;
}

#include<bits/stdc++.h>
#define yuasfy7189dyjkwqafyi7ahakjsfgb872314fg7b8qgcrg4cagujbyrf67b4gs using
#define tqgyfduasyufasfg3814aw678f432yasd6f7g32yaw687f34fae6fuaafdsag86yg5234fg9gy namespace
#define ydwuiygtqffafyut34gf78bgasg67f8afgf4367ffasd1tygf9q67f342avs6f7ys6adfyvuyayhufd std
yuasfy7189dyjkwqafyi7ahakjsfgb872314fg7b8qgcrg4cagujbyrf67b4gs tqgyfduasyufasfg3814aw678f432yasd6f7g32yaw687f34fae6fuaafdsag86yg5234fg9gy ydwuiygtqffafyut34gf78bgasg67f8afgf4367ffasd1tygf9q67f342avs6f7ys6adfyvuyayhufd;
#define ttttyuasfgqewrgvbdfchjafashgfjq3uy long
#define fgtdhsajfgejykfgyagsdgyfkafgrewgfasdtasasdjksdaf 0
#define uyftgdasjkhfgadsyufsdahjfgasdfjasdfgdsa return
#define eqtryuiwqfgghashjfgcasdfcasfdsafgjherwbvuyfa long
#define qweyuasdfhdsabjfghafdsgfvuqewrgftvfuayjdsfasdjyttfgvdsaghsda main
#define rtqyuwefgdsjhavfasgjvfgrweqfybvdastfvbyutqwerkyufttyu <%
#define tftasdjferwgqjtyufkgyudgvafadsyjfsdaykjfgasdyfgvsdjjfh %>
ttttyuasfgqewrgvbdfchjafashgfjq3uy eqtryuiwqfgghashjfgcasdfcasfdsafgjherwbvuyfa fagydfdeykajfviureqgfhdasvtyujdasfgdsatyfgjhtfgajsdfdasdffasdgrgsfdfads;
eqtryuiwqfgghashjfgcasdfcasfdsafgjherwbvuyfa ttttyuasfgqewrgvbdfchjafashgfjq3uy hgfyuafujerwbgvfkdgasfjhasdtyweqrfygdtsajfgsdajyhuftgdsajtyudasfsadthftgsadhfjdasgfjasdgfhjghjsahdfghasdf;
#define gfydufkafyergwfjhdasfsdakjufytgdsafgutdsamfygsdatytugfhfvadsdsa scanf
#define fygiuasdkjfywerqyfiqwegfvyhsafvaskhjfgdshkjfgsdafygkdsagftgdsayjf printf
int qweyuasdfhdsabjfghafdsgfvuqewrgftvfuayjdsfasdjyttfgvdsaghsda()
rtqyuwefgdsjhavfasgjvfgrweqfybvdastfvbyutqwerkyufttyu
gfydufkafyergwfjhdasfsdakjufytgdsafgutdsamfygsdatytugfhfvadsdsa("%lld%lld",&fagydfdeykajfviureqgfhdasvtyujdasfgdsatyfgjhtfgajsdfdasdffasdgrgsfdfads,&hgfyuafujerwbgvfkdgasfjhasdtyweqrfygdtsajfgsdajyhuftgdsajtyudasfsadthftgsadhfjdasgfjasdgfhjghjsahdfghasdf);
fygiuasdkjfywerqyfiqwegfvyhsafvaskhjfgdshkjfgsdafygkdsagftgdsayjf("%lld",fagydfdeykajfviureqgfhdasvtyujdasfgdsatyfgjhtfgajsdfdasdffasdgrgsfdfads+hgfyuafujerwbgvfkdgasfjhasdtyweqrfygdtsajfgsdajyhuftgdsajtyudasfsadthftgsadhfjdasgfjasdgfhjghjsahdfghasdf);
uyftgdasjkhfgadsyufsdahjfgasdfjasdfgdsa fgtdhsajfgejykfgyagsdgyfkafgrewgfasdtasasdjksdaf;
tftasdjferwgqjtyufkgyudgvafadsyjfsdaykjfgasdyfgvsdjjfh

写成这样了，居然还能跑。

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

# include <bits/stdc++.h>
using namespace std;
int main() {
    long long a, b;
    cin >> a >> b;
    cout << a + b;
    return 0;
}
很难的一道题

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

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

我知道你们都看这里，所以我只放答案啦~~~

Code:

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

#include <bits/stdc++.h> using namespace std;
int main() {
long long a, b;
cin >> a >> b;
cout << a + b;
return 0;
}
兄啊，这事给若至做的罢（恼

这道题，别看难度1，其实非常难，需要用到结构体

#include <bits/stdc++.h>
using namespace std;
long long a;
long long b;
long long c;
struct AB{
	void in(){
		cin>>a>>b;
	}
	void js(){
		c+=a;
		c+=b;
	}
	void out(){
		cout<<c;
	}
};
int main(){
	AB ab;
	ab.in(); // 输入 
	ab.js(); // 计算 
	ab.out(); // 输出 
	return 0;
}

`

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