C语言解答

int main(int argc, char* argv[])
{
	int a,b,c,i;
	double sum,count;
	while(~scanf("%d%d%d",&a,&b,&c))
	{
		sum=0;count=0;
	   sum+=a*(a+1)/2.0;
	  for(i=1;i<=b;i++)
        count+=i*i;
	   sum+=count;
      count=0;
      for(i=1;i<=c;i++)
        count+=1.0/i;
	  printf("%.2lf\n",sum+count);
	}
	return 0;
}

C++解答

#include<iostream>
#include<iomanip>
#include<fstream>
#include<string> 
#include<climits>
#include<cctype>
#include<cmath>
#include<cstring>
using namespace std;

int main()
{
 //ifstream cin("aaa.txt");
  int i,j,n,m,k,p,len,l,count;
  
 int c;
  float a,b,sum;
  sum=0;
  string s;
  cin>>a>>b>>c;
  sum+=1.0*(1+a)*a/2+1.0*b*(b+1)*(2*b+1)/6;
 
  for(i=1;i<=c;i++)
  {
	  sum+=1.0/i;
  }
  cout<<fixed<<setprecision(2)<<sum;
   cout<<endl;
  

  	return 0;

}