水上游戏
给定一个空杯子，这个杯子必须装满水，任务是找出杯子在任何时候都能容纳的最大水量。
给定条件：
The program takes an input N denoting the number of steps.
Each step consists of two inputs T and X where T is the flag condition denoting whether the X ml of water has to be poured or drinked based on below conditions:
1. if T = 0, Pour X ml (millilitres) of water in the glass
2. if T = 1, Drink X ml of water from the glass
例子：
Input: N = 4,
0 1
0 1
0 1
1 3
Output: 3
Explanation:
The glass initially has 0 ml of water.
The maximum value is obtained
after the first 3 operations.
Input: N = 2
1 15
1 24
Output: 39
方法该方法可以简单地说为每当 T 为 0 时将体积相加并找到该体积的最大值。
执行：
C++
// C++ implementation of the approach
#include
using namespace std;
// returns the largest volume
int largest_volume(int n, int *t, int *x)
{
// arbitrarily large value
int minValue = 100000000;
// stores the maximum
int maxValue = 0;
// Current Volume of the glass
int c = 0;
for (int i = 0; i < n; i++)
{
// if 1st operation is performed
if (t[i] == 0)
{
// increment with current x
c += x[i];
// take current max
maxValue = max(maxValue, c);
}
// if 2nd operation is performed
else
{
// decrement with current x
c -= x[i];
// take current min
minValue = min(minValue, c);
}
}
// returns the largest difference
return maxValue - minValue;
}
// Driver code
int main()
{
int n = 4;
int t[4] = {0};
int x[4] = {0};
t[0] = 0;
x[0] = 1;
t[1] = 0;
x[1] = 1;
t[2] = 0;
x[2] = 1;
t[3] = 1;
x[3] = 3;
// Find the largest volume
int ans = largest_volume(n, t, x);
// Print the largest volume
cout<< ans;
return 0;
}
// This code is contributed by PrinciRaj1992 Java
// Java implementation of the approach
class GFG {
// returns the largest volume
static int largest_volume(int n, int[] t, int[] x)
{
// arbitrarily large value
int min = 100000000;
// stores the maximum
int max = 0;
// Current Volume of the glass
int c = 0;
for (int i = 0; i < n; i++) {
// if 1st operation is performed
if (t[i] == 0) {
// increment with current x
c += x[i];
// take current max
max = Math.max(max, c);
}
// if 2nd operation is performed
else {
// decrement with current x
c -= x[i];
// take current min
min = Math.min(min, c);
}
}
// returns the largest difference
return max - min;
}
// Driver code
public static void main(String[] args)
{
int n = 4;
int[] t = new int[4];
int[] x = new int[4];
t[0] = 0;
x[0] = 1;
t[1] = 0;
x[1] = 1;
t[2] = 0;
x[2] = 1;
t[3] = 1;
x[3] = 3;
// Find the largest volume
int ans = largest_volume(n, t, x);
// Print the largest volume
System.out.println(ans);
}
}Python3
# Python3 implementation of the approach
# Returns the largest volume
def largest_volume(n, t, x):
# Arbitrarily large value
Min = 100000000
# Stores the maximum
Max = 0
# Current Volume of the glass
c = 0
for i in range(0, n):
# If 1st operation is performed
if t[i] == 0:
# Increment with current x
c += x[i]
# Take current max
Max = max(Max, c)
# If 2nd operation is performed
else:
# Decrement with current x
c -= x[i]
# Take current min
Min = min(Min, c)
# Returns the largest difference
return Max - Min
# Driver code
if __name__ == "__main__":
n = 4
t = [0, 0, 0, 1]
x = [1, 1, 1, 3]
# Find the largest volume
ans = largest_volume(n, t, x)
# Print the largest volume
print(ans)
# This code is contributed by Rituraj JainC#
// C# implementation of the approach
using System;
class GFG
{
// returns the largest volume
static int largest_volume(int n, int[] t, int[] x)
{
// arbitrarily large value
int min = 100000000;
// stores the maximum
int max = 0;
// Current Volume of the glass
int c = 0;
for (int i = 0; i < n; i++)
{
// if 1st operation is performed
if (t[i] == 0)
{
// increment with current x
c += x[i];
// take current max
max = Math.Max(max, c);
}
// if 2nd operation is performed
else
{
// decrement with current x
c -= x[i];
// take current min
min = Math.Min(min, c);
}
}
// returns the largest difference
return max - min;
}
// Driver code
public static void Main()
{
int n = 4;
int[] t = new int[4];
int[] x = new int[4];
t[0] = 0;
x[0] = 1;
t[1] = 0;
x[1] = 1;
t[2] = 0;
x[2] = 1;
t[3] = 1;
x[3] = 3;
// Find the largest volume
int ans = largest_volume(n, t, x);
// Print the largest volume
Console.WriteLine(ans);
}
}
// This code is contributed by
// tufan_gupta2000输出：
3