给定两个整数R1、R2表示两个球员的得分， B1、B2 分别表示他们面对的球，任务是找到可以得到的R1 * R2的最小值，使得R1和R2可以减少M运行满足条件R1 ≥ B1和R2 ≥ B2 。
例子：
Input: R1 = 21, B1 = 10, R2 = 13, B2 = 11, M = 3
Output: 220
Explanation: Minimum product that can be obtained is by decreasing R1 by 1 and R2 by 2, i.e. (21 – 1) x (13 – 2) = 220.
Input: R1 = 7, B1 = 6, R2 = 9, B1 = 9, M = 4
Output: 54
方法：可以通过将数量完全减少到它们的极限来获得最小乘积。将R1减小到其极限B1 ，然后尽可能减小R2 （不超过M ）。类似地，将R2减少到最多 B2，然后尽可能减少R2 （不超过M ）。两种情况下得到的最小乘积就是要求的答案。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Utility function to find the minimum
// product of R1 and R2 possible
int minProductUtil(int R1, int B1, int R2,
int B2, int M)
{
int x = min(R1-B1, M);
M -= x;
// Reaching to its limit
R1 -= x;
// If M is remaining
if (M > 0)
{
int y = min(R2 - B2, M);
M -= y;
R2 -= y;
}
return R1 * R2;
}
// Function to find the minimum
// product of R1 and R2
int minProduct(int R1, int B1, int R2, int B2, int M)
{
// Case 1 - R1 reduces first
int res1 = minProductUtil(R1, B1, R2, B2, M);
// Case 2 - R2 reduces first
int res2 = minProductUtil(R2, B2, R1, B1, M);
return min(res1, res2);
}
// Driver Code
int main()
{
// Given Input
int R1 = 21, B1 = 10, R2 = 13, B2 = 11, M = 3;
// Function Call
cout << (minProduct(R1, B1, R2, B2, M));
return 0;
}
// This code is contributed by maddler Java
// Java program for the above approach
import java.io.*;
class GFG{
// Utility function to find the minimum
// product of R1 and R2 possible
static int minProductUtil(int R1, int B1, int R2,
int B2, int M)
{
int x = Math.min(R1-B1, M);
M -= x;
// Reaching to its limit
R1 -= x;
// If M is remaining
if (M > 0)
{
int y = Math.min(R2 - B2, M);
M -= y;
R2 -= y;
}
return R1 * R2;
}
// Function to find the minimum
// product of R1 and R2
static int minProduct(int R1, int B1, int R2, int B2, int M)
{
// Case 1 - R1 reduces first
int res1 = minProductUtil(R1, B1, R2, B2, M);
// Case 2 - R2 reduces first
int res2 = minProductUtil(R2, B2, R1, B1, M);
return Math.min(res1, res2);
}
// Driver Code
public static void main (String[] args)
{
// Given Input
int R1 = 21, B1 = 10, R2 = 13, B2 = 11, M = 3;
// Function Call
System.out.print((minProduct(R1, B1, R2, B2, M)));
}
}
// This code is contributed by shivanisinghss2110Python3
# Python program for the above approach
# Utility function to find the minimum
# product of R1 and R2 possible
def minProductUtil(R1, B1, R2, B2, M):
x = min(R1-B1, M)
M -= x
# Reaching to its limit
R1 -= x
# If M is remaining
if M > 0:
y = min(R2-B2, M)
M -= y
R2 -= y
return R1 * R2
# Function to find the minimum
# product of R1 and R2
def minProduct(R1, B1, R2, B2, M):
# Case 1 - R1 reduces first
res1 = minProductUtil(R1, B1, R2, B2, M)
# case 2 - R2 reduces first
res2 = minProductUtil(R2, B2, R1, B1, M)
return min(res1, res2)
# Driver Code
if __name__ == '__main__':
# Given Input
R1 = 21; B1 = 10; R2 = 13; B2 = 11; M = 3
# Function Call
print(minProduct(R1, B1, R2, B2, M))C#
// C# program for the above approach
using System;
class GFG{
// Utility function to find the minimum
// product of R1 and R2 possible
static int minProductUtil(int R1, int B1, int R2,
int B2, int M)
{
int x = Math.Min(R1-B1, M);
M -= x;
// Reaching to its limit
R1 -= x;
// If M is remaining
if (M > 0)
{
int y = Math.Min(R2 - B2, M);
M -= y;
R2 -= y;
}
return R1 * R2;
}
// Function to find the minimum
// product of R1 and R2
static int minProduct(int R1, int B1, int R2, int B2, int M)
{
// Case 1 - R1 reduces first
int res1 = minProductUtil(R1, B1, R2, B2, M);
// Case 2 - R2 reduces first
int res2 = minProductUtil(R2, B2, R1, B1, M);
return Math.Min(res1, res2);
}
// Driver Code
public static void Main (String[] args)
{
// Given Input
int R1 = 21, B1 = 10, R2 = 13, B2 = 11, M = 3;
// Function Call
Console.Write((minProduct(R1, B1, R2, B2, M)));
}
}
// This code is contributed by shivanisinghss2110Javascript
输出：
220时间复杂度： O(1)
辅助空间： O(1)