程序检查物体是否处于平衡状态

给定一个2D数组a [] [] ，其中每行由N个向量坐标组成，它们的形式为所应用物体的(x _i ，y _i ，z _i ) 。如果身体处于平衡状态，请打印“是”。否则，打印“ NO ”。

例子：

Input: a[][] = {{4, 1, 7}, {-2, 4, -1}, {1, -5, -3}}
Output: NO

Input: a[][] = {{3, -1, 7}, {-5, 2, -4}, {2, -1, -3}}
Output: YES

方法：请按照以下步骤解决问题：

• 在所有矢量坐标上应用矢量加法。
• 如果向量相加的结果为(0，0，0) ，则物体处于平衡状态。因此，打印“是”。
• 否则，打印“ NO ”。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to calculate vector addition
void vectorAddition(int a[][3], int N)
{
 
    // Sum1 to store sum of xi
    // Sum2 to store sum of yi
    // Sum3 to store sum of zi
    int sum1 = 0, sum2 = 0, sum3 = 0;
 
    for (int i = 0; i < N; i++) {
        sum1 += a[i][0];
        sum2 += a[i][1];
        sum3 += a[i][2];
    }
 
    // If the sum1, sum2 and sum3
    // are all equal to zero
    if (sum1 == 0 && sum2 == 0
        && sum3 == 0) {
 
        // Body is in
        // equilibrium
        cout << "YES";
    }
    else {
 
        // Body is not in
        // equilibrium
        cout << "NO";
    }
}
 
// Driver Code
int main()
{
    int N = 3;
 
    int a[N][3]
        = { { 3, -1, 7 },
            { -5, 2, -4 },
            { 2, -1, -3 } };
 
    vectorAddition(a, N);
 
    return 0;
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG{
 
// Function to calculate
// vector addition
static void vectorAddition(int a[][],
                           int N)
{
  // Sum1 to store sum of xi
  // Sum2 to store sum of yi
  // Sum3 to store sum of zi
  int sum1 = 0, sum2 = 0, sum3 = 0;
 
  for (int i = 0; i < N; i++)
  {
    sum1 += a[i][0];
    sum2 += a[i][1];
    sum3 += a[i][2];
  }
 
  // If the sum1, sum2 and sum3
  // are all equal to zero
  if (sum1 == 0 && sum2 == 0 &&
      sum3 == 0)
  {
    // Body is in
    // equilibrium
    System.out.print("YES");
  }
  else
  {
    // Body is not in
    // equilibrium
    System.out.print("NO");
  }
}
 
// Driver Code
public static void main(String[] args)
{
  int N = 3;
  int a[][] = {{3, -1, 7},
               {-5, 2, -4},
               {2, -1, -3}};
  vectorAddition(a, N);
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 program to implement
# the above approach
 
# Function to calculate vector addition
def vectorAddition(a, N):
     
    # Sum1 to store sum of xi
    # Sum2 to store sum of yi
    # Sum3 to store sum of zi
    sum1 = 0
    sum2 = 0
    sum3 = 0
 
    for i in range(N):
        sum1 += a[i][0]
        sum2 += a[i][1]
        sum3 += a[i][2]
 
    # If the sum1, sum2 and sum3
    # are all equal to zero
    if (sum1 == 0 and sum2 == 0 and
        sum3 == 0):
 
        # Body is in
        # equilibrium
        print("YES")
 
    else:
 
        # Body is not in
        # equilibrium
        print("NO")
 
# Driver Code
if __name__ == '__main__':
     
    N = 3
    a = [ [ 3, -1, 7 ],
          [ -5, 2, -4 ],
          [ 2, -1, -3 ] ]
 
    vectorAddition(a, N)
 
# This code is contributed by mohit kumar 29

C#

// C# Program to implement
// the above approach
using System;
class GFG{
 
// Function to calculate
// vector addition
static void vectorAddition(int[,]a,
                           int N)
{
  // Sum1 to store sum of xi
  // Sum2 to store sum of yi
  // Sum3 to store sum of zi
  int sum1 = 0, sum2 = 0, sum3 = 0;
 
  for (int i = 0; i < N; i++)
  {
    sum1 += a[i, 0];
    sum2 += a[i, 1];
    sum3 += a[i, 2];
  }
 
  // If the sum1, sum2 and sum3
  // are all equal to zero
  if (sum1 == 0 && sum2 == 0 &&
      sum3 == 0)
  {
    // Body is in
    // equilibrium
    Console.Write("YES");
  }
  else
  {
    // Body is not in
    // equilibrium
    Console.Write("NO");
  }
}
 
// Driver Code
public static void Main(String[] args)
{
  int N = 3;
  int[,]a = {{3, -1, 7},
             {-5, 2, -4},
             {2, -1, -3}};
  vectorAddition(a, N);
}
}
 
// This code is contributed by shikhasingrajput

输出

YES

时间复杂度： O(N)
辅助空间： O(1)