5 solutions
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2
#include<bits/stdc++.h> using namespace std; double n; int a[105]; double q,w,e,r; int main(){ cin>>n; for(int i=0;i<n;i++){ cin>>a[i]; if(0<=a[i]&&a[i]<=18) q++; if(19<=a[i]&&a[i]<=35) w++; if(36<=a[i]&&a[i]<=60) e++; if(a[i]>=61) r++; } q=q/n*100; w=w/n*100; e=e/n*100; r=r/n*100; cout<<fixed<<setprecision(2)<<q<<"%"<<endl; cout<<fixed<<setprecision(2)<<w<<"%"<<endl; cout<<fixed<<setprecision(2)<<e<<"%"<<endl; cout<<fixed<<setprecision(2)<<r<<"%"<<endl; return 0; }
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1
#include <bits/stdc++.h> using namespace std; int r[4],n,age; int main(){ cin>>n; for(int i=0;i<n;i++){ cin>>age; if(age>=0&&age<=18) r[0]++; else if (age>=19&&age<=35) r[1]++; else if (age>=36&&age<=60) r[2]++; else r[3]++; } for(int i=0;i<4;i++) cout<<fixed<<setprecision(2)<<(double)r[i]/n*100<<'%'<<endl; return 0; }
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1
#include <bits/stdc++.h> using namespace std; int r[4]; int main(){ int n , age; cin >> n; for(int i = 0; i < n;i++){ cin >> age; if(age >= 0 && age <= 18) r[0]++; else if (age >= 19 && age <= 35) r[1]++; else if (age >= 36 && age <= 60) r[2]++; else r[3]++; } for(int i = 0; i < 4; ++i) cout << fixed << setprecision(2) << (double)r[i] / n * 100 << '%' << endl; return 0; }
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1
#include<bits/stdc++.h> using namespace std;
int main(){ double a,b,c1=0,c2=0,c3=0,c4=0; double g1,g2,g3,g4; cin>>a; for(int i=1;i<=a;i++){ cin>>b; if(b<=18){ c1++; }if(b>=19&&b<=35){ c2++; }if(b>=36&&b<=60){ c3++; }else if(b>=61){ c4++; } } g1=c1/a100; g2=c2/a100; g3=c3/a100; g4=c4/a100; printf("%.2f",g1); cout<<"%"<<endl; printf("%.2f",g2); cout<<"%"<<endl; printf("%.2f",g3); cout<<"%"<<endl; printf("%.2f",g4); cout<<"%"<<endl; }//0-18、19-35、36-60、61
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0
#include<bits/stdc++.h>
using namespace std; int main(){ int n; double sum0=0,sum1=0,sum2=0,sum3=0;
cin>>n; const int N=n; int a[N]; for(int i=0;i<N;i++){ cin>>a[i];
} for(int k=0;k<N;k++){ if(a[k]>=0&&a[k]<=18){ sum0+=1; } else if(a[k]>=19&&a[k]<=35){ sum1+=1; } else if(a[k]>=36&&a[k]<=60){ sum2+=1; } else{ sum3+=1; } } cout<<fixed<<setprecision(2)<<sum0/n100<<"%"<<endl;; cout<<fixed<<setprecision(2)<<sum1/n100<<"%"<<endl;; cout<<fixed<<setprecision(2)<<sum2/n100<<"%"<<endl;; cout<<fixed<<setprecision(2)<<sum3/n100<<"%"<<endl;;
return 0;
}
- 1
Information
- ID
- 592
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 2
- Tags
- # Submissions
- 38
- Accepted
- 25
- Uploaded By