求给定规则生成的矩阵元素之和
给定三个整数 A 、 B 和 R ，任务是找出由给定规则生成的矩阵的所有元素的和:
- 第一行将包含单个元素 A ，其余元素为 0 。
- 下一行将包含两个元素，全部为 (A + B) ，其余为 0s 。
- 第三行将包含 (A + B + B) 三次，其余为 0s 以此类推。
- 矩阵将只包含 R 行。
比如如果 A = 5 、 B = 3 和 R = 3 那么矩阵就是: 5 0 0 8 8 0 11 11 11 例:
输入: A = 5，B = 3，R = 3 输出:54 5+8+8+11+11 = 54 输入: A = 7，B = 56，R = 1 输出: 7
方法:初始化总和= 0 并且对于每个 1 ≤ i ≤ R 更新总和=总和+ (i * A) 。每次迭代更新后 A = A + B 。打印最后的总和。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to return the required sum
int sum(int A, int B, int R)
{
// To store the sum
int sum = 0;
// For every row
for (int i = 1; i <= R; i++) {
// Update the sum as A appears i number
// of times in the current row
sum = sum + (i * A);
// Update A for the next row
A = A + B;
}
// Return the sum
return sum;
}
// Driver code
int main()
{
int A = 5, B = 3, R = 3;
cout << sum(A, B, R);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// JAVA implementation of the approach
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
// Function to return the required sum
static int sum(int A, int B, int R)
{
// To store the sum
int sum = 0;
// For every row
for (int i = 1; i <= R; i++)
{
// Update the sum as A appears i number
// of times in the current row
sum = sum + (i * A);
// Update A for the next row
A = A + B;
}
// Return the sum
return sum;
}
// Driver code
public static void main (String[] args)
throws java.lang.Exception
{
int A = 5, B = 3, R = 3;
System.out.print(sum(A, B, R));
}
}
// This code is contributed by nidhiva
Python 3
# Python3 implementation of the approach
# Function to return the required ssum
def Sum(A, B, R):
# To store the ssum
ssum = 0
# For every row
for i in range(1, R + 1):
# Update the ssum as A appears i number
# of times in the current row
ssum = ssum + (i * A)
# Update A for the next row
A = A + B
# Return the ssum
return ssum
# Driver code
A, B, R = 5, 3, 3
print(Sum(A, B, R))
# This code is contributed by Mohit Kumar
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the required sum
static int sum(int A, int B, int R)
{
// To store the sum
int sum = 0;
// For every row
for (int i = 1; i <= R; i++)
{
// Update the sum as A appears i number
// of times in the current row
sum = sum + (i * A);
// Update A for the next row
A = A + B;
}
// Return the sum
return sum;
}
// Driver code
public static void Main ()
{
int A = 5, B = 3, R = 3;
Console.Write(sum(A, B, R));
}
}
// This code is contributed by anuj_67..
java 描述语言
<script>
// JAVA SCRIPT implementation of the approach
// Function to return the required sum
function sum( A, B, R)
{
// To store the sum
let sum = 0;
// For every row
for (let i = 1; i <= R; i++)
{
// Update the sum as A appears i number
// of times in the current row
sum = sum + (i * A);
// Update A for the next row
A = A + B;
}
// Return the sum
return sum;
}
// Driver code
let A = 5, B = 3, R = 3;
document.write(sum(A, B, R));
//contributed by bobby
</script>
Output:
54
时间复杂度: O(R)
辅助空间: O(1)
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