7 条题解

  • 1
    @ 2025-2-6 15:04:30

    #include<bits/stdc++.h> using namespace std; #define LL long long const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; long long ans,n; int main() { cin >> n; ans = n % 4; if ( ans == 1 ) cout << 1; else if ( ans == 2 ) cout << n+1; else if ( ans == 3 ) cout << 0; else cout << n; return 0; }

    • 1
      @ 2023-4-21 21:32:48
      #include<bits/stdc++.h>
      #define fi first
      #define se second
      #define INF 0x3f3f3f3f
      #define ll long long
      #define ld long double
      #define mem(ar,num) memset(ar,num,sizeof(ar))
      #define me(ar) memset(ar,0,sizeof(ar))
      #define lowbit(x) (x&(-x))
      #define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
      #define lcm(a,b) ((a)*(b)/(__gcd((a),(b))))
      using namespace std;
      const int N = 1e3 + 10;
      const int mod = 1e9 + 7;
      ll n;
      int main() {
          ll ans = 0;
          cin >> n;
          if(!(n & 1))
              n++, ans = n;
          ll num = (n + 1ll) / 2ll;
          if(num & 1)
              ans ^= 1;
          cout << ans;
          return 0;
      }
      
      • 0
        @ 2026-4-19 11:23:36

        #include<bits/stdc++.h> #define fi first #define se second #define INF 0x3f3f3f3f #define ll long long #define ld long double #define mem(ar,num) memset(ar,num,sizeof(ar)) #define me(ar) memset(ar,0,sizeof(ar)) #define lowbit(x) (x&(-x)) #define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); #define lcm(a,b) ((a)*(b)/(__gcd((a),(b)))) using namespace std; const int N = 1e3 + 10; const int mod = 1e9 + 7; ll n; int main() { ll ans = 0; cin >> n; if(!(n & 1)) n++, ans = n; ll num = (n + 1ll) / 2ll; if(num & 1) ans ^= 1; cout << ans; return 0; }

        • 0
          @ 2025-7-17 15:47:16
          #include<bits/stdc++.h>
          using namespace std;
          long long n,cnt;
          int main(){
          	cin>>n;
          	if(n<=3)for(int i=1;i<=n;i++)cnt^=i;
          	else for(long long i=n-n%4;i<=n;i++)cnt^=i;
          	cout<<cnt;
          	return 0;
          }
          
          

          主要难在找规律防超时

          • -1
            @ 2023-4-29 11:30:57
            /*60代码
            #pragma GCC optimize(3)
            #include<bits/stdc++.h>
            using namespace std;
            long long ans = 1;
            int main() {
            long long n;
            cin >> n;
            for (int i = 2; i <= n; i++) ans = (ans ^ i);
            cout << ans << endl;
            }
            */
            
            /*80代码
            #include<bits/stdc++.h>
            using namespace std;
            
            long long ans = 1;
            int main() {
            int n;
            cin >> n;
            while (n != 0) {		
            if (n & 1 == 0) {
            cout << 1;
            return 0;
            }
            //		cout << n << endl;
            n = (n >> 1);
            }
            cout << 0;
            return 0;
            }
            */
            
            //100AC
            #include<bits/stdc++.h>
            using namespace std;
            
            int main() {
            long long n;
            cin >> n;
            int k = n % 4;
            if (k == 1) cout << 1;
            else if (k == 2) cout << n + 1;
            else if (k == 3) cout << 0;
            else cout << n;
            return 0;
            }
            
            • @ 2023-12-3 15:48:23

              什么原理?

          • -1
            @ 2023-4-29 11:27:56
            /*找规律:
            1. 1
            2. 3
            3. 0
            4 .4
            
            5. 1
            6. 7
            7 .0
            8. 9
            ...
            */
            #include<iostream>
            using namespace std;
            long long n;
            signed main(){
                cin>>n;
                switch (n%4)
                {
                case 0:
                    cout<<n;
                    break;
                case 1:
                    cout<<1;
                    break;
                case 2:
                    cout<<n+1;
                    break;
                default:
                    cout<<0;
                }
                return 0;
            }
            
            • -1
              @ 2023-4-29 11:24:12
              #include<bits/stdc++.h>
              using namespace std;
              #define int long long
              int n;
              signed main(){
              	cin>>n;
              	if(n%4==3) cout<<0<<endl;
              	if(n%4==0) cout<<n<<endl;
              	if(n%4==1) cout<<1<<endl;
              	if(n%4==2) cout<<((n/4+1)*4)-1<<endl;
                  return 0;
              }
              
              • 1

              信息

              ID
              2555
              时间
              1000ms
              内存
              256MiB
              难度
              7
              标签
              递交数
              643
              已通过
              128
              上传者