VT06 - Tính trung bình cộng của mảng
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Gửi bài giải
Điểm:
1,00 (OI)
Giới hạn thời gian:
1.0s
Giới hạn bộ nhớ:
256M
Tác giả:
Dạng bài
Ngôn ngữ cho phép
C, C#, C++, Go, Java, Pascal, Perl, PHP, PyPy, Python, Ruby, Rust, Scratch, Swift
Bạn hãy viết chương trình tính trung bình cộng của tất cả các phần tử lẻ trong mảng 1 chiều các số nguyên ~A~ có ~n~ phần tử.
Input
- Dòng 1, một số nguyên dương ~n~ là số lượng phần tử của mảng ~A~
- Dòng 2, ~n~ số nguyên tương ứng là các phần tử của mảng ~A~
Biết rằng
- ~1 \le n \le 10^3~
- ~|A_{ij}| \le 10^6~
Output
- Trung bình cộng các phần tử lẻ của mảng ~A~ làm tròn tới chữ số thập phân thứ 4
Sample
Input #1
5
1 2 3 4 6
Output #1
2.0000
Bình luận
⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⢿⠿⠿⠿⠿⠿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠏⠓⠀⠀⠀⠀⠀⠀⠀⠀⠈⠙⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿ ⣧⣛⣟⣻⣛⢿⠿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡟⢆⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢌⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿ ⣿⣿⣿⣷⣿⣯⣿⣶⣯⣾⣽⢯⡿⣿⣿⣿⣿⣿⣿⣿⣿⡙⠀⠀⠀⠀⠀⣀⣀⣀⣤⣀⡀⠀⢘⣾⣿⣿⣿⣿⣿⣿⣿⣿⣿ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣷⣿⣿⣿⣻⣿⣿⣷⣿⣾⣶⣠⢠⣾⣿⣿⣿⣿⣿⣿⣷⡐⣾⣿⣿⣿⣿⣿⣿⣿⣿⡿ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡟⣯⠉⢻⣿⣷⡌⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣯⣷⣮⡙⣿⣿⠘⣿⡟⣾⣿⣿⣦⠙⣿⡿⣼⣿⣿⣿⣿⣿⣿⣿⣿⣯ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⢣⣸⣿⠀⠜⢿⣿⣿⣿⣷⡷⠋⠁⢻⣿⣿⣿⣿⣿⣿⣿⣿⣟ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⠿⢿⠿⠿⣫⡇⠰⢥⡀⠙⠛⠋⠁⠀⢀⣠⣾⣿⣿⣿⣿⣿⣿⣿⣿⣯ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣷⣤⣄⣤⣶⣿⣀⣀⣉⣹⣿⣿⣾⣷⣾⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡷ ⢻⢻⣿⡽⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡟⣿⣿⣿⣿⡟⠈⠛⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣟ ⠀⠀⠚⢽⣽⢻⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡼⣿⣷⣿⣿⣿⣷⣶⣶⣿⠟⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣯ ⠀⠀⠀⠀⢟⣣⡝⢿⣿⣻⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⢻⣿⣿⣿⣿⣷⣿⡿⠃⢸⣼⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣧ ⠀⠀⠀⠀⠀⣚⠹⠈⣿⣳⣟⡾⣿⣿⣿⣿⣿⣿⣿⣿⣿⡈⣿⣿⣿⣿⣋⣁⣄⣶⣿⣟⣿⣻⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡷ ⠀⠀⠀⠀⠀⠠⡁⠀⣟⡲⣭⢟⣷⣻⢿⣿⣿⣿⣿⣿⣿⡅⣿⣿⣿⣿⣿⣿⣿⣿⢻⡼⣞⢷⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣟ ⠀⠀⠀⠀⢀⠡⠐⠀⣏⠖⡭⣞⣳⢯⣿⣻⣿⢿⣿⣿⣿⡇⣿⣿⣿⣿⡿⣿⡹⣎⢷⣹⣞⣿⡉⢹⣿⣿⣿⣿⣿⣿⣿⣿⣯ ⢌⡡⢂⠅⠢⢌⠡⢠⡟⡼⢱⢎⣳⢻⡾⣽⣻⣿⣿⣿⣿⣷⢹⣿⣿⣯⢗⡧⣟⡼⣯⢷⣻⣿⣯⠙⣿⣿⣿⣿⣿⣿⣿⣿⡷ ⢦⡱⣊⠼⣑⢮⡱⢠⣿⣌⡷⣮⣝⡷⣻⢷⣻⣽⣿⣿⣿⣿⢸⣿⣿⣞⣯⣞⢷⣫⢿⣽⣻⣿⣿⡐⣿⣿⣿⣿⣿⣿⣿⣿⡿ ⢧⠳⡭⢞⡽⢮⣗⢸⡿⣾⣽⣳⣿⣽⣯⣿⢯⣿⣾⣿⣿⣿⣼⣿⣿⣟⣮⣟⣯⣽⣻⢾⡽⣿⣿⡿⢹⣿⣿⣿⣿⣿⣿⣿⣿ ⠎⣱⠹⡈⡾⢱⢇⢸⡿⣰⢇⠿⡾⣿⣏⣿⣿⣿⣿⣿⣷⣿⣿⣿⣿⣿⡿⣾⢷⣾⡹⣏⡿⣿⡿⠀⣿⣿⣿⣿⣿⣿⣿⣿⣿ ⡰⣤⣤⣵⣾⢿⣏⢸⣷⣶⣮⣷⣳⣾⣼⣶⣟⣮⣷⣿⣿⣿⣿⣿⣿⡿⣽⣻⣞⣧⢿⣽⣻⣿⡟⢰⣿⣿⣿⣿⣿⣿⣿⣿⣿ ⣿⣿⣻⣟⣯⣛⢮⢸⣿⡻⣽⢻⡿⣿⣻⡿⣿⣿⣿⣿⡟⣿⣿⣿⣿⣿⣷⣻⢾⣭⣟⣾⣳⣿⡇⣿⣿⣿⣿⣿⣿⣿⣿⣿⣻ ⣿⣿⣽⣟⡶⣯⣟⣸⣿⡱⣏⣷⣻⢷⣯⣟⣿⣿⣿⣿⣇⢻⣿⣿⣿⣿⣷⣿⣻⢾⣽⣾⣿⢋⢸⣿⣿⣿⣿⣿⣿⣿⣿⣿⣷ ⣿⣿⣿⣯⢿⣷⣿⢸⣿⣳⢟⣾⣽⣟⣾⣽⣻⣿⣿⣿⣿⣾⣿⣿⣿⣿⣿⢾⣻⣯⢷⣿⣾⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿ ⣿⣿⣿⣿⣟⣯⣿⢺⣿⣽⣻⣞⣷⣻⣿⣾⣿⣿⣿⣿⣿⠟⣽⣿⣿⣿⣿⣿⣻⣞⣯⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿ ⡿⢻⢟⣿⣟⡿⣿⣾⡿⠿⠿⣿⡿⣿⠿⢿⣿⡿⣿⢟⣃⣾⣿⣿⣿⣿⣿⣿⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿ ⠀⠀⡀⠀⠀⡀⡙⡛⢷⣦⣤⣄⣀⣀⣠⢀⡤⣤⡄⢫⣾⣧⣔⣿⣿⣿⣿⣽⣫⣿⣿⣻⣿⣷⣷⣻⣾⣽⣾⣿⣿⣿⣿⣿⣿ ⣭⡷⣼⣝⣳⣽⣳⣟⣯⣿⣼⣿⣿⣿⣿⣿⣿⠟⢬⣿⣿⣿⣾⣿⡽⣿⣿⣿⣿⣿⡷⣭⣿⣿⣿⣿⣿⣿⣿⣟⣿⣛⣛⣛⣻ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠫⣱⣿⣿⣿⣿⣿⣯⣹⣿⣿⣿⣿⣿⣇⢳⣻⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⠗⣰⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠋⣴⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿ ⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡟⢄⣾⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⢿⣿⣿⣿⣿⡿⣿⢿⣿⣿⣿ ⣿⣿⣿⡿⢿⠿⡿⢿⢿⡻⣿⠕⣊⣾⣿⣯⣿⣭⢿⣻⢻⢯⢿⣹⣿⣿⣿⣏⢷⣙⣾⣿⣿⣿⢸⢣⣿⣿⣿⣽⣯⣿⣾⣿⣿ ⣧⣤⣤⣿⢿⣿⣻⣟⡿⣿⢧⢛⣧⣿⡿⣧⣟⡼⣟⣧⣟⡿⣧⢿⣼⣿⣿⡿⣼⣻⣼⣿⣿⣇⡿⣜⣻⢿⠿⣟⡻⣿⢿⣿⣿ ⣿⣹⣟⠾⣏⠶⣫⢿⡽⣏⢮⣟⣿⡿⣽⣳⢯⣿⣽⣾⣯⣿⣿⣿⣿⣿⣿⣿⣿⣷⣿⣿⣿⣿⣿⣯⣷⣫⠞⣥⠳⡌⢎⡍⢆ ⢦⠱⡌⠒⣌⠲⣳⣯⢟⣼⣻⣾⣿⣛⡷⣯⣿⣿⣿⣿⣿⣿⣿⣿⠟⢿⣾⣿⣿⣿⣿⡿⣾⣿⣿⣿⣷⣯⡟⣶⡙⡜⢢⠜⢢ ⢌⠣⡘⣁⠦⣽⣟⣧⢿⣾⣿⡿⣲⢯⢿⣽⣻⣿⣿⣿⣿⢟⠫⠁⣘⣮⢿⣿⣿⣿⣿⠀⢻⣿⣿⣿⣿⣳⡟⢦⠓⣌⠣⡘⠄ ⡌⠱⠐⢬⣿⣿⣽⣾⣿⣿⢟⣨⣵⣿⣾⠿⠟⢫⣍⡉⠈⠀⣠⢞⣵⣿⣿⣿⣿⠏⠃⠀⡑⢺⣿⣟⢯⡗⠻⡌⢃⠆⡡⠘⡄ ⠬⡑⡈⠘⣿⣿⣿⣿⣿⡿⠟⠛⠋⠉⢸⣦⣯⣁⢡⣐⣦⣿⣾⡿⠿⠛⢉⠻⢡⣁⢠⡄⣦⢡⡹⣎⠒⡌⠱⡈⢆⡘⢄⠣⡐ ⢣⡔⢡⢻⡿⠿⠿⢛⠋⣔⠣⡐⢄⠂⡌⡙⢿⡿⢿⠟⢋⠙⠃⠐⡀⠄⣾⣤⣽⣗⣺⣗⣿⣿⣷⡿⠆⡌⡱⢌⠲⡌⢎⣱⡱
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include <bits/stdc++.h>
using namespace std; using ll = long long;
int main() { ios::syncwithstdio(false); cin.tie(nullptr);
}
include <bits/stdc++.h>
using namespace std; using ll = long long;
int main() { ios::syncwithstdio(false); cin.tie(nullptr);
}
include <bits/stdc++.h>
using namespace std; using ll = long long;
int main() { ios::syncwithstdio(false); cin.tie(nullptr);
}
chao mn
include <bits/stdc++.h>
using namespace std; long long a; long long s=0; long long c=0; long long b[1000001]; long double tbc; int main() { cin>>a; for(long long i=0;i<a;i++){ cin>>b[i]; } for(long long i=0;i<a;i++){ if(b[i]%2!=0){ s=s+b[i]; c++; } } tbc=(double)s/c; cout<<fixed<<setprecision(4)<<tbc; }
hi mn
include <bits/stdc++.h>
using namespace std; int main() { int n; scanf("%d", &n); int a[n]; long long sum = 0; // Dùng long long để tránh tràn số nếu tổng lớn int cnt = 0;
}
FULL AC CHO AE
include <bits/stdc++.h>
using namespace std; long long a[1001]; long long i,h,s,m; double k,n,l; int main() { cin>>n; k=0; l=0; for (i=1;i<=n;i++) { cin>>a[i]; } for (i=1;i<=n;i++) { if(a[i]%2!=0) { l=l+1; } } for (i=1;i<=n;i++) { if(a[i]%2!=0) { k=k+a[i]; } } cout<<fixed<<setprecision(4)<<k/l; } chúc anh em làm tốt nhé
FULL AC nha các ae ( tui không cần vote miễn sao bạn thích là đc:)
include <bits/stdc++.h>
using namespace std; long long a[100],n,i,d; float trungbinh,tong; int main() { { cin>>n; tong=0; d=0; for(i=0;i<=n-1;i++) cin>>a[i]; for(i=0;i<=n-1;i++) { if(a[i]%2!=0) { tong=tong+a[i]; d++; } } trungbinh=tong/d; cout<<fixed<<setprecision(4)<<trungbinh; } return 0; }
include<bits/stdc++.h>
using namespace std; int i,n,s,d,tbc; int a[100]; int main() { cin>>n; for(i=1;i<=n;i++) cin>>a[i]; for(i=1;i<=n;i++) if(a[i]%2!=0) {s=s+a[i]; d=d+1;} cout<<setprecision(4)<<fixed<<float(s)/d; return 0; }
Full ac nha!
include<bits/stdc++.h>
using namespace std; int i,n,s,d,tbc; int a[100]; int main() { cin>>n; for(i=1;i<=n;i++) cin>>a[i]; for(i=1;i<=n;i++) if(a[i]%2!=0) {s=s+a[i]; d=d+1;} cout<<setprecision(4)<<fixed<<float(s)/d; return 0; }
include<bits/stdc++.h>
using namespace std; int i,n,s,d,tbc; int a[100]; int main() { cin>>n; for(i=1;i<=n;i++) cin>>a[i]; for(i=1;i<=n;i++) if(a[i]%2!=0) {s=s+a[i]; d=d+1;} cout<<setprecision(4)<<fixed<<float(s)/d; return 0; }
hello ae
có ai làm đc bài này ko chỉ mình với các bạn
include <bits/stdc++.h>
using namespace std; long long a[1001]; long long n; long double chuso = 0.0; long double tong = 0.0; int main() { cin>>n; for(int i = 1;i <= n ;i ++) { cin>>a[i]; } for(int i = 1 ; i <= n ;i++) { if(a[i] % 2 != 0) { chuso = chuso + 1; } } for(int i = 1 ; i <=n ;i ++) { if(a[i] % 2 != 0) { tong = tong + a[i]; } } cout<<fixed<<setprecision(4)<<tong/chuso; return 0; }
include <iostream>
include <iomanip>
using namespace std; long long a[100],n,i,dem; float trungbinh , t; int main() { cin >> n; t = 0; dem =0; for (i=0;i<=n-1;i++) cin >> a[i]; for (i=0;i<=n-1;i++) { if (a[i] % 2 != 0) { t = t + a[i]; dem ++; } } trungbinh = t / dem; cout << fixed << setprecision(4) << trungbinh << endl;
return 0; }
có khi Bạn chưa Xem kỹ output , output bảo Trung bình cộng các phần tử lẻ của mảng nhé
cảm ơn nhé a bạn!
Các bạn nhớ xử lý trường hợp số lẻ là số âm nhé
đây là tính các số lẻ trong mảng hay là các vị trí lẻ ạ? em làm mãi không ra
tinhs soo ler nhes banj
code c++ thì có nên dùng chương trình con ở bài này ko
Bình luận này đã bị ẩn vì có quá nhiều phản ứng tiêu cực. Nhấn để xem.
Chắc Là Bạn chưa Xem kỹ output , output bảo Trung bình cộng các phần tử lẻ của mảng nhé