4 条题解

  • 1
    @ 2026-1-13 22:44:53
    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    typedef pair<int, int> pii;
    #define endl '\n'
    #define mem(a, b) memset(a, b, sizeof(a))
    #define rep(i, a, b) for (int i = a; i <= b; ++i)
    #define rrep(i, a, b) for (int i = a; i >= b; --i)
    #define FOR(i, a, b) for (int i = a; i < b; ++i)
    #define pb push_back
    #define MP make_pair
    #define all(v) v.begin(), v.end()
    #define clr(v) v.clear()
    #define isEven(x) !(x & 1)
    #define yes cout<<"YES"<<endl
    #define no cout<<"NO"<<endl
    #define cinv(x) cin >> x
    #define coutv(x) cout << x
    int main() {
    	ios::sync_with_stdio ( false );
    	cin.tie ( 0 );
    	long long n;
    	cinv ( n );
    	vector<int>a ( n );
    	for ( long long i = n / 2; i * i > 1; --i ) while ( n % i == 0 ) {
    			coutv ( i ) ;
    			coutv ( endl );
    			return 0;
    		}
    
    	return 0;
    }
    
    • 1
      @ 2025-12-9 19:04:06
      #include<bits/stdc++.h>
      using namespace std;
      int main(){
          int n;
          cin>>n;
          for(int i=2;i<=n;i++) 
          	if(n%i==0) { 
            	    cout<<n/i<<endl;
            	    break;
          	}
          return 0;
      }
      
      • -2
        @ 2025-12-4 20:29:26

        #include #include using namespace std;

        int main() { long long n; cin >> n;

        // 从2开始寻找n的第一个质因数(即较小的那个质数)
        for (long long i = 2; i * i <= n; i++) {
            if (n % i == 0) {
                // 找到了较小的质因数i,较大的质数就是n/i
                cout << n / i << endl;
                return 0;
            }
        }
        
        return 0;
        

        }

        • -2
          @ 2025-12-4 20:29:14

          #include #include using namespace std;

          int main() { long long n; cin >> n;

          // 从2开始寻找n的第一个质因数(即较小的那个质数)
          for (long long i = 2; i * i <= n; i++) {
              if (n % i == 0) {
                  // 找到了较小的质因数i,较大的质数就是n/i
                  cout << n / i << endl;
                  return 0;
              }
          }
          
          return 0;
          

          }

          • 1

          信息

          ID
          6941
          时间
          1000ms
          内存
          128MiB
          难度
          3
          标签
          递交数
          28
          已通过
          19
          上传者