2 条题解

  • 0
    @ 2025-7-14 8:34:06

    正解:

    #include<bits/stdc++.h> using namespace std; int fastPowMod(int a, int b, int m) { int result = 1; a = a % m; while (b > 0) { if (b % 2 == 1) { result = (result * a) % m; } a = (a * a) % m; b = b / 2; } return result; } int main() { int a, b; cin >> a >> b; int mod = fastPowMod(a, b, 7); string weekdays[7] = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"}; cout << weekdays[mod] << endl; return 0; }

    • 0
      @ 2025-6-5 20:46:22

      #include

      #include

      using namespace std;

      // 快速幂计算 a^b mod m

      int fastPowMod(int a, int b, int m) { int result = 1;

      a = a % m; // 先取模,防止a过大
      
      while (b > 0)
      
      {
          
          if (b % 2 == 1)
          {
             
              result = (result * a) % m;
          
          }
          
          a = (a * a) % m;
          
          b = b / 2;
          
      }
      
      return result;
      

      }

      int main()

      {

      int a, b;
      
      cin >> a >> b;
      
      // 计算 a^b mod 7
      
      int mod = fastPowMod(a, b, 7);
      
      string weekdays[7] = {"Sunday",
      
      "Monday", "Tuesday", "Wednesday", 
      
      "Thursday", "Friday", "Saturday"};
      
      cout << weekdays[mod] << endl;
      
      return 0;
      

      }

      • 1

      信息

      ID
      87
      时间
      1000ms
      内存
      256MiB
      难度
      4
      标签
      递交数
      60
      已通过
      26
      上传者