7 条题解

  • 1
    @ 2023-3-11 21:23:41
    #include<iostream>
    #include<iomanip>
    using namespace std;
    int main()
    {
        int n,tmp,a1=0,a2=0,a3=0,a4=0;
        cin>>n;
        for(int i=0;i<n;i++)
        {
            cin>>tmp;
            if(tmp<=18)a1+=1;
            else if(19<=tmp&&tmp<=35)a2+=1;
            else if(36<=tmp&&tmp<=60)a3+=1;
            else if(tmp>=61)a4+=1;
        }
        cout<<fixed<<setprecision(2);
        cout<<a1*100.0/n<<'%'<<endl;
        cout<<a2*100.0/n<<'%'<<endl;
        cout<<a3*100.0/n<<'%'<<endl;
        cout<<a4*100.0/n<<'%'<<endl;
        return 0;
    }
    
    
    • 0
      @ 2024-6-11 21:00:45

      #include<iostream> #include<iomanip> using namespace std; int main() { double n,tmp,a1=0,a2=0,a3=0,a4=0; cin>>n; for(int i=0;i<n;i++) { cin>>tmp; if(tmp<=18)a1+=1; else if(19<=tmp&&tmp<=35)a2+=1; else if(36<=tmp&&tmp<=60)a3+=1; else if(tmp>=61)a4+=1; } cout<<fixed<<setprecision(2); cout<<a1100.0/n<<'%'<<endl; cout<<a2100.0/n<<'%'<<endl; cout<<a3100.0/n<<'%'<<endl; cout<<a4100.0/n<<'%'<<endl; }

      • 0
        @ 2024-5-18 15:57:07
        #include <bits/stdc++.h>
        #define maxn 300000
        using namespace std;
        typedef long long ll;
        int n,a[maxn],h1,h2,h3,h4;
        double l;
        int main(){
        	cin>>n;
        	for(int i=1;i<=n;i++){
        		cin>>a[i];
        		if(a[i]<=18) h1++;
        		else if(a[i]<=35) h2++;
        		else if(a[i]<=60) h3++;
        		else h4++;
        	}
        	printf("%.2lf\%\n%.2lf\%\n%.2lf\%\n%.2lf\%",h1*100.0/n,h2*100.0/n,h3*100.0/n,h4*100.0/n);
        }
        
        • 0
          #include <bits/stdc++.h>
          using namespace std;
          const int N=1e7+10;
          const int INF=0x3f3f3f3f;
          int a[N],n,z1,z2,z3,z4; 
          int main()
          {
          	cin>>n;
          	for(int i=1;i<=n;i++)
          	{
          		cin>>a[i];
          		if(a[i]>=0 && a[i]<=18)
          		{
          			z1++;
          		}
          		else if(a[i]>=19 && a[i]<=35)
          		{
          			z2++;
          		}
          		else if(a[i]>=36 && a[i]<=60)
          		{
          			z3++;
          		}
          		else if(a[i]>=61)
          		{
          			z4++;
          		}
          	}
          	cout<<fixed<<setprecision(2)<<z1*100.0/n<<"%\n";
          	cout<<fixed<<setprecision(2)<<z2*100.0/n<<"%\n";
          	cout<<fixed<<setprecision(2)<<z3*100.0/n<<"%\n";
          	cout<<fixed<<setprecision(2)<<z4*100.0/n<<"%\n";
          	return 0;
          }
          
          • 0
            @ 2023-4-4 22:26:48
            #include<iostream>
            using namespace std;
            int n,a,b,c,d;
            void out(int x){
            	printf("%.2lf",100.0*x/n);
            	cout<<"%\n";
            }
            int main(){
            	int x;
            	cin>>n;
            	for(int i=1;i<=n;i++){
            		cin>>x;
            		if(0<=x&&x<=18)a++;
            		else if(x<=35)b++;
            		else if(x<=60)c++;
            		else d++;
            	}
            	out(a);out(b);out(c);out(d);
            	return 0;
            }
            
            • 0
              @ 2023-2-19 15:00:51

              #include <bits/stdc++.h>

              using namespace std;

              int a[1005];

              int main(){

              int n,x,ans1 = 0,ans2 = 0,ans3 = 0,ans4 = 0;

              cin>>n;

              for(int i =1;i<=n;i++){

              cin>>x;

              if(0 <= x && x <= 18) ans1++;

              else if(x<=35) ans2++;

              else if(x<=60) ans3++;

              else ans4++;

              }

              cout<<fixed<<setprecision(2);

              cout<<1.0ans1 / n100<<"%"<<endl;

              cout<<1.0ans2 / n100<<"%"<<endl;

              cout<<1.0ans3 / n100<<"%"<<endl;

              cout<<1.0ans4 / n100<<"%"<<endl;

              }

              • 0
                @ 2022-8-12 20:43:30
                #include <bits/stdc++.h>
                
                using namespace std;
                
                //0-18、19-35、36-60、more than 61
                
                int p_q[101];
                
                int main(void)
                {
                	int n;
                	
                	cin >> n;
                	
                	float q0, q19, q36, q61;
                	
                	
                	for (int i = 0; i <= n - 1; i++)
                	{
                		cin >> p_q[i];
                	}
                	
                	for (int i = 0; i <= n - 1; i++)
                	{
                		if (p_q[i] <= 18 && p_q >= 0)
                		{
                			q0++;
                		}
                		
                		if (p_q[i] <= 35 && p_q[i] >= 19)
                		{
                			q19++;
                		}
                		
                		if (p_q[i] <= 60 && p_q[i] >= 36)
                		{
                			q36++;
                		}
                		
                		if (p_q[i] >= 61)
                		{
                			q61++;
                		}
                	}
                	
                	printf("%0.2f%\n", q0 / n * 100);
                	printf("%0.2f%\n", q19 / n * 100);
                	printf("%0.2f%\n", q36 / n * 100);
                	printf("%0.2f%%", q61 / n * 100);
                	
                	return 0;
                }
                
                • 1

                信息

                ID
                1006
                时间
                1000ms
                内存
                128MiB
                难度
                5
                标签
                递交数
                382
                已通过
                156
                上传者