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| Wrong answer. Help pls!! C++ | TIU_Sarexer | 1991. The battle near the swamp | 9 Mar 2022 01:34 | 4 |
#include <iostream>
using namespace std;
int main() {
int n, k, *arr,raz,y=0,ost=0;
cin >> n >> k;
arr = new int[n];
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
for (int i = 0; i < n; i++) {
raz = 5 - arr[i];
if (raz >= 0) {
y = y + raz;
}
else {
y = y;
ost = ost + raz * -1;
}
}
cout << ost << " " << y;
return 0;
} > raz = 5 - arr[i];
Why "5" ? ahahah I'm stupid :D Thanks) the task could be solved by shorter way
// 2022-03-08.cpp : This file contains the 'main' function. Program execution begins and ends there.
//
#include <iostream>
using namespace std;
int main()
{
int n, k;
cin >> n >> k;
int droid = 0, bum = 0;
int tempBum;
for (int i = 0; i < n; i++) {
cin >> tempBum;
if (tempBum > k) {
bum += tempBum - k;
}
if (k > tempBum) {
droid += k - tempBum;
}
}
cout << bum << " " << droid;
}
|
| easy translation | Mattiev Jamol | 1991. The battle near the swamp | 24 Jan 2022 14:41 | 4 |
if(a[i]<k){
ans+=k-a[i];
}
else{
ans1+=a[i]-k;
} I think that this problem is easy enough to try to make the solution as efficient as possible.
My consideration notes:
1. You can make it without using arrays, because the local information that you get is enough to generate the answer.
2. I think that it is worth trying to make the speaking variable names. It will certainly require additional time, but I think, it should pay off in the future. Any good book (about Pascal for example) will tell about it Ho do they get 0.001 seconds? |
| help me, find mistake, 5 test | buyolitsez | 1991. The battle near the swamp | 13 Nov 2018 21:04 | 1 |
#include <bits/stdc++.h>
using namespace std;
#define int long long
signed main() {
ios_base::sync_with_stdio(false);
cout.tie(0);
cin.tie(0);
int n,k;
cin >> n >> k;
int dontused=0,dontdie=0;
int f;
for(int i=0;i<n;i++)
{
cin >> f;
if(k>f)
{
dontdie+=k-f;
}
if(k<f)
{
dontused+=f-k;
}
}
cout << dontused << " " << dontdie;
return 0;
} |
| what mistake? | Evgenii | 1991. The battle near the swamp | 13 Jun 2017 18:03 | 2 |
program pr;
Var
n,k,i,s,q:integer;
a:array[1..100000] of integer;
b:array[1..100000] of integer;
Begin
q:=0;
read(n,k);
for i:=1 to n do
readln(a[i]);
for i:=1 to n do
b[i]:=k-a[i];
for i:=1 to n do begin
if b[i]<0 then s:=abs(b[i]);
if b[i]>=0 then q:=q+b[i] ;
end;
write(s,' ',q)
End. |
| 1991. Битва у болота // please,need help! where is the mistake& | kodirmullaboyev | 1991. The battle near the swamp | 25 Jan 2016 12:34 | 1 |
import java.util.Scanner;
/**
* Created by Oxygen! on 25.01.2016.
*/
public class BitvaUBolota1991Var2 {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int x=0,y=0;
int n=sc.nextInt();
int k=sc.nextInt();
int a[]=new int[n];
for(int i=0; i<n; i++) {
a[i]=sc.nextInt();
if(a[i]<=0) {
y+=k-a[i];
} else {
x+=a[i]-k;
}
System.out.println(x+" "+y);
}
}
} |
| admin plz help!! | Shubham Murarka | 1991. The battle near the swamp | 20 Sep 2015 18:24 | 2 |
what is the range of "n" ?? I am not the admin, but I will help :)
From 1 to 10000. |
| I have AC in Java 1.7 | Axmadjon | 1991. The battle near the swamp | 6 May 2014 19:09 | 1 |
|
| To admins | Silent | 1991. The battle near the swamp | 4 Nov 2013 14:48 | 5 |
я знаю что она решается, у меня AC. я говорю про то, что на Go 1.1 простой способ считывания входных данных работает некорректно, и прошу разобраться - это бага fmt.Scanf или что-то с форматом test #3. Try to read integers in the first string using format "%v %v\n" instead of "%v %v". It's not evident, but... it works! If solution reads the first string using fmt.Scanf("%v %v\n",&n,&k) then it gets AC. Thanks, Sandro (USU)! |
