| Бодолт илгээх | Бүх бодолтууд | Шилдэг бодолтууд | Жагсаалт руу буцах |
ABR0035A - Максимум 3 |
x, y, z бодит тоонууд нэг мөрөнд, зайгаар тусгаарлагдан өгөгдөнө.
Input
x, y, z бодит тоонууд нэг мөрөнд, зайгаар тусгаарлагдан өгөгдөнө.
Output
max(x+y+z, xyz) тоог таслалаас хойш нэг оронгийн нарийвчлалтайгаар гаргана.
Example
Input: 1 1 1 Output: 3.0
| Нэмсэн: | sw40 |
| Огноо: | 2007-10-20 |
| Хугацааны хязгаарлалт: | 1s |
| Эх кодын хэмжээний хязгаарлалт: | 50000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Програмчлалын хэлүүд: | Бүгд дараах хэлүүдээс бусад: ASM32-GCC MAWK BC C-CLANG NCSHARP CPP14 CPP14-CLANG COBOL COFFEE D-CLANG D-DMD DART ELIXIR ERL FANTOM FORTH GOSU GRV JS-RHINO JS-MONKEY JULIA KTLN NIM NODEJS OBJC OBJC-CLANG OCT PERL6 PICO PROLOG PYPY PYPY3 PY_NBC R RACKET RUST CHICKEN SQLITE SWIFT UNLAMBDA VB.NET |
| Эх сурвалж: | Абрамов С. А. |
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2024-02-19 05:03:43
⠀⠀⠀⠀⠀⠀⢀⣴⡻⠋⠁⠀⠀⠀⠐⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠻⣶⡀⠀⡿⠦⣤⣤⡈⠳⡙⢦⡀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⣠⠛⢹⣅⣠⠤⠒⠒⠚⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠘⣷⡀⡇⠀⠀⠙⠉⠛⠛⠀⠙⣄⠀⠀⠀⠀ ⠀⠀⠀⠀⡼⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠺⠿⠃⠀⠀⠀⠀⠀⠀⠀⠀⠸⢦⠀⠀⠀ ⠀⠀⠀⡼⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣀⣀⣤⡤⠤⠤⠤⠤⠤⠤⢤⣀⣀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⢧⠀⠀ ⠀⠀⡼⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣠⠴⠚⣻⡟⠁⢰⠛⡇⠀⠀⠀⠀⠀⠀⠀⠀⠈⠹⣦⠀⠀⠀⠀⠀⠀⠀⠀⠈⡇⠀ ⠀⢰⠃⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⡴⠞⠉⠀⣀⡾⢽⠗⢲⠏⠘⠋⠉⢛⠉⠉⠉⠉⠑⠲⣄⡀⠈⢧⡀⠀⠀⠀⠀⠀⠀⠀⢸⡀ ⠀⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣤⠞⠉⣼⠤⠖⡻⠃⠀⢸⠀⡜⠀⠀⠀⠀⡘⠀⠀⠀⠀⠀⡄⠈⢳⡀⠀⢳⡀⠀⠀⠀⠀⠀⠀⠀⠁ ⢸⠁⠀⠀⠀⠀⠀⠀⠀⠀⢀⣴⣏⡤⠒⢩⠇⠀⡴⠁⠤⢀⣘⣸⡀⠀⢀⣧⢀⣿⣸⡀⠀⠀⢀⡇⠀⢀⠙⣆⠀⢻⡀⠀⠀⠀⠀⠀⠀⢀ ⠈⠀⠀⠀⠀⠀⠀⠀⢀⣴⠿⠋⢡⠀⢠⠋⣀⠞⠀⠀⠀⠀⢸⡏⠀⠀⠀⢇⢸⠙⣿⠀⠀⠀⡸⡇⠀⣾⠀⠈⢷⡀⢳⡀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⢀⣴⠿⡇⠀⠀⡇⠀⡏⢠⣏⠀⠀⠀⠀⠀⠘⠀⠀⠀⠀⢸⡇⠀⣿⠀⠀⢠⠃⡇⣰⠀⡇⠀⢸⢳⡈⣧⠀⠀⠀⠀⠀⢠ ⠀⠀⠀⠀⠀⢀⡼⠋⠀⡇⠀⢸⣇⡸⠀⠈⠛⢿⣶⣤⣄⣐⠠⠤⡀⠀⠀⠘⠁⠀⠏⠀⢀⠎⠀⣿⠷⣄⣣⠀⡎⠀⠹⣼⡄⠀⠀⠀⠀⡏ ⠀⠀⠀⠀⢠⠞⠀⠀⠀⠸⡄⢸⢻⡇⠀⠀⠀⢸⣇⠈⠉⠛⣿⡿⠇⠀⠀⠀⠀⠀⡇⣠⠋⠀⠘⠁⠀⠀⠙⢲⠇⠀⠀⢹⣧⠀⠀⠀⢠⠁ ⠀⠀⠀⢠⣿⠀⠀⠀⠀⠀⣿⡼⢼⠇⠀⠀⠀⠘⠿⣶⣴⡾⠋⠀⠀⠀⠀⠀⠀⢸⠟⠁⣷⣦⣤⠀⠀⠀⢰⢻⠀⠀⣰⠀⣿⠀⠀⢀⡞⠀ ⠀⠀⣰⣿⣿⠀⠀⠀⠀⠀⢸⣿⣼⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⣿⠿⠷⣶⣦⣤⣼⡀⠀⡇⠀⣿⡀⠀⣼⠁⠀ ⠀⢠⣿⣿⣿⠀⠀⠀⠀⠀⢸⣿⢻⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢿⣤⡀⢀⣼⡿⠙⠀⣸⠃⢸⣿⠀⣼⠇⠀⠀ ⠀⣼⣿⣿⣿⡆⠀⠀⠀⠀⠀⣿⠸⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠛⠻⠛⠋⠀⡇⡰⠟⢀⡟⣿⣠⠏⠀⠀⠀ ⢠⣿⣿⣿⣿⣇⠀⠀⠀⠀⠀⣿⢦⡁⠀⠀⠀⠀⠀⢀⣄⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣼⠞⢁⣧⠞⢀⣿⡏⠀⠀⠀⠀ ⣾⣿⣿⣿⣿⣿⣄⠀⠀⠀⠀⣿⠀⣷⡄⠀⠀⠀⠀⢸⠛⠛⠲⢤⣀⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣼⣷⢄⡾⢃⡴⣼⢻⠃⠀⠀⠀⠀ ⣿⣿⣿⣿⣿⣿⠏⠀⠀⠀⠀⣿⡄⢿⠙⣦⡀⠀⠀⠈⠦⣀⡀⠀⢻⣿⠇⠀⠀⠀⠀⠀⠀⠀⠀⢀⣾⠟⣡⠏⢀⡞⣽⠃⣾⠀⠀⠀⠀⠀ ⠻⣿⣿⣿⣿⡟⠀⠀⠀⠀⢀⣿⣷⣿⢠⠇⠙⢦⡀⠀⠀⠀⠉⠉⠉⠀⠀⠀⠀⠀⠀⠀⠀⢀⡴⢿⣿⡄⠏⣠⣿⠞⠁⠀⣿⠀⠀⠀⠀⠀ ⠀⠹⣿⣿⡟⠀⠀⠀⠀⠀⢸⣿⣿⡏⣸⠀⠀⠀⠉⠢⣄⣀⣀⣀⣀⣀⣀⣠⣤⣤⠴⠒⠚⢉⡴⠋⢸⣠⡾⠛⠁⠀⠀⠀⣿⠀⠀⠀⠀⠀ ⡀⠀⢹⡿⠁⠀⠀⠀⠀⠀⣾⣿⣿⣧⡏⠀⠀⠀⠀⠀⠀⠀⢀⣠⣿⣥⣾⣿⡟⠀⠀⢀⡴⠋⠀⢀⣿⠋⠀⠀⣀⣀⣠⣤⡿⠀⠀⠀⠀⠀ ⠙⢦⣴⠃⠀⠀⠀⠀⠀⢸⠋⠉⠻⣿⡀⠀⠀⠀⠀⢀⡤⠞⠋⠁⠀⠀⢈⣹⠇⢀⡴⠋⠀⠀⠀⣼⣷⣶⣾⣿⣿⣿⣿⣿⠃⠀⠀⠀⠀⠀ ⠀⠀⣙⠀⠀⠀⠀⠀⠀⣏⣀⣀⡀⠙⠿⣦⣀⣤⢶⠋⠀⠀⣠⠶⠶⠚⠉⣹⣰⠞⠀⠀⠀⠀⢰⠏⠉⠉⠙⢻⣿⣿⠟⠁⠀⠀⠀⠀⠀⠀ |
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2020-09-21 15:45:46
import math x , y , z = input().split() x = float(x) y = float(y) z = float(z) sum = x + y + x product = x * y * z if sum > product : print("{:.1f}".format(sum)) else : print("{:.1f}".format(product)) |
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2017-10-18 08:45:47
#include<stdio.h> int main() { float x, y, z, sum, product; scanf("%f %f %f", &x, &y, &z); sum=x+y+z; product=x*y*z; if(sum>product) printf("%.1f", sum); else printf("%.1f", product); return 0; } Миний хувьд ингээд л бодчихлоо. Бодлогын өгүүлбэрийг зөв ойлгох нь үгүй ээ мөн чухал шүү. Өгүүлбэрээ сайн ойлгоогүйгээс болж нэг хоёр буруу код бичлээ |
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2017-04-04 15:03:54
c++ deer bodlogo bodoh geheer hurwvvleltiin aldaa geed bh ym? |
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2012-12-14 03:56:24 M.Enkhbileg
#include <iostream> #include <iomanip.h> #include <stdlib.h> using namespace std; main() { double x, y, z, amount1, amount2; cin >> x >> y >> z; amount1 = x + y + z; amount2 = x*y*z; if(amount1 > amount2) cout << resetiosflags(ios::showpoint)<< setiosflags(ios::fixed)<<setprecision(1)<< amount1 <<endl; else cout << resetiosflags(ios::showpoint)<< setiosflags(ios::fixed)<<setprecision(1)<< amount2 <<endl; system("PAUSE"); return 0; } Last edit: 2012-12-14 03:57:05 |
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2012-10-25 14:35:08 sw12
#include <stdio.h> #include <stdlib.h> #include<math.h> main() { double i, b,sum=0; for(int i=3;i<=103;i=i+2) { b=(1+1/3); sum+=1/(b+(1/(i+2)));} printf("%lf", sum); system("pause"); return 0; } ene bodlogo zov uu |
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2012-10-05 09:33:43 Tso .
hi |
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2012-04-20 10:13:44 oidowjunai
#include "stdio.h" #include "math.h" #include "stdlib.h" int main(){ double x, y, z, max, c1, c2; scanf("%lf%lf%lf", &x, &y, &z); c1 = (x+y+z); c2 = (x*y*z); if(c1>c2){ max = c1; }else{ max = c2; } printf("%.1lf", max); system("pause"); } |
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2011-10-05 13:42:43 hohohoho
#include "stdio.h" #include "math.h" #include "stdlib.h" int main(){ double x, y, z, max, c1, c2; scanf("%lf%lf%lf", &x, &y, &z); c1 = (x+y+z); c2 = (x*y*z); if(c1>c2){ max = c1; }else{ max = c2; } printf("%.1lf", max); system("pause"); } |
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2011-01-04 08:57:47 khasa
#include<stdlib.h> #include<stdio.h> int main() { float a,b,; scanf("%f",&a,&b); printf("%.1f\n%.1f\n%.1f",a-b,a*b,a/b); return 0; |
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