7 solutions

  • 3
    @ 2024-4-1 11:16:48
    #include <iostream>
    using namespace std;
    
    int main() {
    	long long lkb, hzy;
    	cin >> lkb >> hzy;
    	cout << lkb + hzy;
    	return 0;
    }
    
    • 2
      @ 2025-1-9 19:20:45
      #include<bits/stdc++.h>
      using namespace std;
      #define int long long
      const int maxn=1e5+5;
      const int mod=1e9+7;
      int a,b;
      signed main() {
      	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
      	cin>>a>>b;
      	cout<<a+b;
      	return 0;
      }
      
      • 2
        @ 2024-3-5 21:55:45
        #include <iostream>
        using namespace std;
        
        int main() {
        	long long lkb, hzy;
        	cin >> lkb >> hzy;
        	cout << lkb + hzy;
        	return 0;
        }
        
        • 1
          @ 2023-10-15 20:39:29
          #include<stdio.h>
          int main(){long long a,b;scanf("%d%d",&a,&b);printf("%d %d",a,b);}
          
          • 0
            @ 2024-9-6 9:23:03

            #include using namespace std;

            int main() { long long ts, th; cin >> ts >> th; cout << ts + th; return 0; }

            • 0
              @ 2023-6-1 22:15:01
              #include <iostream>
              using namespace std;
              
              int main() {
              	long long lkb, hzy;
              	cin >> lkb >> hzy;
              	cout << lkb + hzy;
              	return 0;
              }
              
              • -4
                @ 2024-8-13 11:37:17

                很明显,新手可能并不知道如何使用加法符号,所以我们可以通过高精度来实现

                
                #include<bits/stdc++.h>
                using namespace std;
                int a[250],b[250],c[250];
                int main(){
                  string A,B;
                  cin>>A>>B;
                  int len=max(A.length(),B.length());
                  for(int i=A.length()-1,j=1;i>=0;i--,j++)a[j]=A[i]-'0';
                  for(int i=B.length()-1,j=1;i>=0;i--,j++)b[j]=B[i]-'0';
                  for(int i=1;i<=len;i++){
                    c[i]+=a[i]+b[i];
                    c[i+1]=c[i]/10;
                    c[i]%=10;
                  }
                  if(c[len+1])len++;
                  for(int i=len;i>=1;i--)cout<<c[i];
                  return 0;
                }
                
                //纯恶意,无娱乐
                
                • 1

                Information

                ID
                6762
                Time
                1000ms
                Memory
                128MiB
                Difficulty
                1
                Tags
                # Submissions
                366
                Accepted
                80
                Uploaded By