油箱变空的天数
原文:https://www . geesforgeks . org/number-days-tank-will-变空/
给定一个容量为 C 升的油箱，它在起动时完全充满。每天水箱都装满 1 升水，如果溢出，多余的水会被倒掉。现在，在第一天，我会拿出几升水喝。我们需要找出油箱第一次变空的日期。
示例:
Input : Capacity = 5
l = 2
Output : 4
At the start of 1st day, water in tank = 5
and at the end of the 1st day = (5 - 1) = 4
At the start of 2nd day, water in tank = 4 + 2 = 6
but tank capacity is 5 so water = 5
and at the end of the 2nd day = (5 - 2) = 3
At the start of 3rd day, water in tank = 3 + 2 = 5
and at the end of the 3rd day = (5 - 3) = 2
At the start of 4th day, water in tank = 2 + 2 = 4
and at the end of the 4th day = (4 - 4) = 0
So final answer will be 4
我们可以看到，在开始的(l + 1)天里，水箱将是满的，因为取出的水比装满的水少。此后，每天水箱中的水将再减少 1 升，在第(l + 1 + i)天(C –( I)(I+1)/2)升水将在饮用前保留。
现在我们需要找到一个最小日(l + 1 + K)，在这个最小日中，即使在将水箱装满 l 升之后，水箱中的水也少于 l，即在第(l+1+K–1)天水箱变空，因此我们的目标是找到最小 K，这样， C–K(K+1)/2<= l
我们可以用二分搜索法解上面的方程，然后(l + K)就是我们的答案。解决方案的总时间复杂度为 0(对数 C)
C++
// C/C++ code to find number of days after which
// tank will become empty
#include <bits/stdc++.h>
using namespace std;
// Utility method to get sum of first n numbers
int getCumulateSum(int n)
{
return (n * (n + 1)) / 2;
}
// Method returns minimum number of days after
// which tank will become empty
int minDaysToEmpty(int C, int l)
{
// if water filling is more than capacity then
// after C days only tank will become empty
if (C <= l)
return C;
// initialize binary search variable
int lo = 0;
int hi = 1e4;
int mid;
// loop until low is less than high
while (lo < hi) {
mid = (lo + hi) / 2;
// if cumulate sum is greater than (C - l)
// then search on left side
if (getCumulateSum(mid) >= (C - l))
hi = mid;
// if (C - l) is more then search on
// right side
else
lo = mid + 1;
}
// final answer will be obtained by adding
// l to binary search result
return (l + lo);
}
// Driver code to test above methods
int main()
{
int C = 5;
int l = 2;
cout << minDaysToEmpty(C, l) << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java code to find number of days after which
// tank will become empty
public class Tank_Empty {
// Utility method to get sum of first n numbers
static int getCumulateSum(int n)
{
return (n * (n + 1)) / 2;
}
// Method returns minimum number of days after
// which tank will become empty
static int minDaysToEmpty(int C, int l)
{
// if water filling is more than capacity then
// after C days only tank will become empty
if (C <= l)
return C;
// initialize binary search variable
int lo = 0;
int hi = (int)1e4;
int mid;
// loop until low is less than high
while (lo < hi) {
mid = (lo + hi) / 2;
// if cumulate sum is greater than (C - l)
// then search on left side
if (getCumulateSum(mid) >= (C - l))
hi = mid;
// if (C - l) is more then search on
// right side
else
lo = mid + 1;
}
// final answer will be obtained by adding
// l to binary search result
return (l + lo);
}
// Driver code to test above methods
public static void main(String args[])
{
int C = 5;
int l = 2;
System.out.println(minDaysToEmpty(C, l));
}
}
// This code is contributed by Sumit Ghosh
Python 3
# Python3 code to find number of days
# after which tank will become empty
# Utility method to get
# sum of first n numbers
def getCumulateSum(n):
return int((n * (n + 1)) / 2)
# Method returns minimum number of days
# after which tank will become empty
def minDaysToEmpty(C, l):
# if water filling is more than
# capacity then after C days only
# tank will become empty
if (C <= l) : return C
# initialize binary search variable
lo, hi = 0, 1e4
# loop until low is less than high
while (lo < hi):
mid = int((lo + hi) / 2)
# if cumulate sum is greater than (C - l)
# then search on left side
if (getCumulateSum(mid) >= (C - l)):
hi = mid
# if (C - l) is more then
# search on right side
else:
lo = mid + 1
# Final answer will be obtained by
# adding l to binary search result
return (l + lo)
# Driver code
C, l = 5, 2
print(minDaysToEmpty(C, l))
# This code is contributed by Smitha Dinesh Semwal.
C
// C# code to find number
// of days after which
// tank will become empty
using System;
class GFG
{
// Utility method to get
// sum of first n numbers
static int getCumulateSum(int n)
{
return (n * (n + 1)) / 2;
}
// Method returns minimum
// number of days after
// which tank will become empty
static int minDaysToEmpty(int C,
int l)
{
// if water filling is more
// than capacity then after
// C days only tank will
// become empty
if (C <= l)
return C;
// initialize binary
// search variable
int lo = 0;
int hi = (int)1e4;
int mid;
// loop until low is
// less than high
while (lo < hi)
{
mid = (lo + hi) / 2;
// if cumulate sum is
// greater than (C - l)
// then search on left side
if (getCumulateSum(mid) >= (C - l))
hi = mid;
// if (C - l) is more then
// search on right side
else
lo = mid + 1;
}
// final answer will be
// obtained by adding
// l to binary search result
return (l + lo);
}
// Driver code
static public void Main ()
{
int C = 5;
int l = 2;
Console.WriteLine(minDaysToEmpty(C, l));
}
}
// This code is contributed by ajit
java 描述语言
<script>
// Javascript code to find number
// of days after which
// tank will become empty
// Utility method to get
// sum of first n numbers
function getCumulateSum(n)
{
return parseInt((n * (n + 1)) / 2, 10);
}
// Method returns minimum
// number of days after
// which tank will become empty
function minDaysToEmpty(C, l)
{
// If water filling is more
// than capacity then after
// C days only tank will
// become empty
if (C <= l)
return C;
// Initialize binary
// search variable
let lo = 0;
let hi = 1e4;
let mid;
// Loop until low is
// less than high
while (lo < hi)
{
mid = parseInt((lo + hi) / 2, 10);
// If cumulate sum is
// greater than (C - l)
// then search on left side
if (getCumulateSum(mid) >= (C - l))
hi = mid;
// If (C - l) is more then
// search on right side
else
lo = mid + 1;
}
// Final answer will be
// obtained by adding
// l to binary search result
return (l + lo);
}
// Driver code
let C = 5;
let l = 2;
document.write(minDaysToEmpty(C, l));
// This code is contributed by rameshtravel07
</script>
输出:
4
替代解: 可以用一个简单的公式进行数学求解:
让我们假设 C>L。假设 d 是 L 次后油箱变空的天数。期间会有 (d-1) 续费、 d 取款。
因此我们需要解这个方程:
所有提款的总和是算术级数的总和，因此:
判别式= 1+8(C-L)>0，因为 C>L.
跳过负根，我们得到如下公式:
因此，最终的答案是:
C++
// C/C++ code to find number of days after which
// tank will become empty
#include <bits/stdc++.h>
using namespace std;
// Method returns minimum number of days after
// which tank will become empty
int minDaysToEmpty(int C, int l)
{
if (l >= C)
return C;
double eq_root = (std::sqrt(1+8*(C-l)) - 1) / 2;
return std::ceil(eq_root) + l;
}
// Driver code to test above methods
int main()
{
cout << minDaysToEmpty(5, 2) << endl;
cout << minDaysToEmpty(6514683, 4965) << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java code to find number of days
// after which tank will become empty
import java.lang.*;
class GFG {
// Method returns minimum number of days
// after which tank will become empty
static int minDaysToEmpty(int C, int l)
{
if (l >= C) return C;
double eq_root = (Math.sqrt(1 + 8 *
(C - l)) - 1) / 2;
return (int)(Math.ceil(eq_root) + l);
}
// Driver code
public static void main(String[] args)
{
System.out.println(minDaysToEmpty(5, 2));
System.out.println(minDaysToEmpty(6514683, 4965));
}
}
// This code is contributed by Smitha Dinesh Semwal.
Python 3
# Python3 code to find number of days
# after which tank will become empty
import math
# Method returns minimum number of days
# after which tank will become empty
def minDaysToEmpty(C, l):
if (l >= C): return C
eq_root = (math.sqrt(1 + 8 * (C - l)) - 1) / 2
return math.ceil(eq_root) + l
# Driver code
print(minDaysToEmpty(5, 2))
print(minDaysToEmpty(6514683, 4965))
# This code is contributed by Smitha Dinesh Semwal.
C
// C# code to find number
// of days after which
// tank will become empty
using System;
class GFG
{
// Method returns minimum
// number of days after
// which tank will become empty
static int minDaysToEmpty(int C,
int l)
{
if (l >= C) return C;
double eq_root = (Math.Sqrt(1 + 8 *
(C - l)) - 1) / 2;
return (int)(Math.Ceiling(eq_root) + l);
}
// Driver code
static public void Main ()
{
Console.WriteLine(minDaysToEmpty(5, 2));
Console.WriteLine(minDaysToEmpty(6514683,
4965));
}
}
// This code is contributed by ajit
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP code to find number
// of days after which
// tank will become empty
// Method returns minimum
// number of days after
// which tank will become empty
function minDaysToEmpty($C, $l)
{
if ($l >= $C)
return $C;
$eq_root = (int)sqrt(1 + 8 *
($C - $l) - 1) / 2;
return ceil($eq_root) + $l;
}
// Driver code
echo minDaysToEmpty(5, 2), "\n";
echo minDaysToEmpty(6514683,
4965), "\n";
// This code is contributed
// by akt_mit
?>
java 描述语言
<script>
// Javascript code to find number
// of days after which
// tank will become empty
// Method returns minimum
// number of days after
// which tank will become empty
function minDaysToEmpty(C, l)
{
if (l >= C) return C;
let eq_root = (Math.sqrt(1 + 8 *
(C - l)) - 1) / 2;
return (Math.ceil(eq_root) + l);
}
// Driver code
document.write(minDaysToEmpty(5, 2) + "</br>");
document.write(minDaysToEmpty(6514683, 4965));
// This code is contributed by suresh07
</script>
输出:
4
8573
感谢 Andrey Khayrutdinov 提出这个解决方案。 本文由 乌塔什·特里维迪 供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用write.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 contribute@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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