8 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
      @ 2024-10-11 22:42:10

      #include <stdio.h> #include <iostream> #include <math.h> #include<iomanip> #include<bits/stdc++.h> using namespace std; int main() { int a,b[100]; float c=0,d=0,e=0,f=0; cin>>a; for(int i=1;i<=a;i++){ cin>>b[i]; } for(int i=1;i<=a;i++){ if(0<=b[i]&&b[i]<=18){ c++; }else{ if(b[i]<=35){ d++; }else{ if(b[i]<=60){ e++; }else{ f++; } } } } c=c/a; d=d/a; e=e/a; f=f/a; cout<<fixed<<setprecision(2)<<c100; cout<<"%"<<endl; cout<<fixed<<setprecision(2)<<d100; cout<<"%"<<endl; cout<<fixed<<setprecision(2)<<e100; cout<<"%"<<endl; cout<<fixed<<setprecision(2)<<f100; cout<<"%"; 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
                  478
                  已通过
                  187
                  上传者