4 条题解

  • 1
    @ 2025-10-2 20:20:49

    这一题的数据很大,longlong都存不下... 记得换成unsign longlong

    下面heky学长写的是高精度算法...

    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h> 
    #include<math.h>
    
    int main(){
    	int n,i;scanf("%d",&n);
    	unsigned long long sum=0;
    	unsigned long long a=1;
    	for(i=0;i<=n;i++){
    		sum+=a;
    		a*=2;
    	}
    	printf("%llu",sum);
    	
        return 0;
    }
    
    • 1
      @ 2024-9-26 13:44:43
      #include<iostream>
      #include<vector>
      using namespace std;
      
      vector<int> mul(vector<int> &A,int b) 
      {
          vector<int> C;
          int t = 0;
          for (int i =0;i<A.size();i++)
          {
              t += A[i]*b;
              C.push_back(t%10);
              t /= 10;
          }
          while(t)
          {
              C.push_back(t%10);
              t /= 10;
          }
          return C;
      }
      vector<int> add(vector<int> &A,vector<int> &B) 
      {
          vector<int> C;
          int t = 0;
          if (A.size() < B.size()) return add(B,A);
          for (int i = 0;i<A.size();i++)
          {
              t += A[i] + B[i];
              C.push_back(t%10);
              t /= 10;
          }
          while (t)
          {
              C.push_back(t%10);
              t /= 10;
          }
          return C;
      }
      int main() 
      {
          int n;
          vector<int> B(1,1);
          cin >> n;
          for (int i = 1;i<=n;i++)
          {
              vector<int> A(1,1);
              for (int j =1;j<=i;j++)
              {
                  A = mul(A,2);
              }
              B = add(B,A);
          }
          for (int i = B.size() - 1;i>= 0;i--) printf("%d",B[i]);
          return 0;
      }
      
      • 0
        @ 2025-10-16 21:03:09
        #include<stdio.h>
        
        int main(){
        	
        	int n;
        	scanf("%d",&n);
        	
        	unsigned long long sum=0;
        	
        	unsigned long long a=1;
        	int i;
        	for(i=0;i<=n;i++){
        		sum+=a;
        		a*=2;
        	} 
        	
        	printf("%llu",sum);
        	
        	return 0;
        	
        }
        
        • 0
          @ 2025-10-5 15:27:25
          #include <iostream>
          using namespace std;
          int main()
          {
              unsigned long long ans = 1, last = 1;
              int n, i;
              cin >> n;
              for (i = 1; i <= n; i++)
              {
                  ans += last * 2;
                  last = last * 2;
              }
          
              cout << ans;
              return 0;
          }
          
          • 1

          信息

          ID
          33
          时间
          1000ms
          内存
          512MiB
          难度
          8
          标签
          递交数
          939
          已通过
          125
          上传者