4 条题解

  • 2
    @ 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
      @ 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;
      }
      
    • 0
      @ 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;
      }
      
      • 0
        @ 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
        8
        标签
        递交数
        577
        已通过
        103
        上传者