连续除法后求数
给定两个数组 div 和 rem ，这两个数组保存了因子和余数的值，任务是找到在被 div 数组的元素连续除之后剩下的余数，这些余数在 rem 数组中。 注:第一次除法的商除以第二个元素，然后所得的商除以第三个元素(与给定的余数相匹配)等等。 举例:
输入: div[] = {3，5，7}，rem[] = {1，3，5} 输出: 85 85 在 3 叶余数 1 的除法上用商 28 28 在 5 叶余数 3 的除法上用商 5 5 在 7 叶余数 5 的除法上用 输入: div[] = {7，9}，rem[] = {2，2} 输出:
进场:
- 将最后一个余数的值存储在一个变量中，比如 num 。
- 从 n-2 向后遍历数组到 0 并将 num 更新为 num = num * div[i] + rem[i]
- 最后打印 num 。
以下是上述方法的实现:
C++
// C++ program to implement above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the number
int findNum(int div[], int rem[], int N)
{
int num = rem[N - 1];
for (int i = N - 2; i >= 0; i--) {
num = num * div[i] + rem[i];
}
return num;
}
// Driver Code
int main()
{
int div[] = { 8, 3 };
int rem[] = { 2, 2 };
int N = sizeof(div) / sizeof(div[0]);
cout << findNum(div, rem, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
public class GFG{
// Function to find the number
static int findNum(int div[], int rem[], int N)
{
int num = rem[N - 1];
for (int i = N - 2; i >= 0; i--) {
num = num * div[i] + rem[i];
}
return num;
}
// Driver Code
public static void main(String []args){
int div[] = { 8, 3 };
int rem[] = { 2, 2 };
int N = div.length;
System.out.println(findNum(div, rem, N));
}
// This code is contributed by ANKITRAI1
}
Python 3
# Python 3 program to implement
# above approach
# Function to find the number
def findNum(div, rem, N):
num = rem[N - 1]
i = N - 2
while(i >= 0):
num = num * div[i] + rem[i]
i -= 1
return num
# Driver Code
if __name__ == '__main__':
div = [8, 3]
rem = [2, 2]
N = len(div)
print(findNum(div, rem, N))
# This code is contributed by
# Surendra_Gangwar
C
// C# implementation of the
// above approach
using System;
class GFG
{
// Function to find the number
static int findNum(int []div,
int []rem, int N)
{
int num = rem[N - 1];
for (int i = N - 2; i >= 0; i--)
{
num = num * div[i] + rem[i];
}
return num;
}
// Driver Code
static public void Main ()
{
int []div = { 8, 3 };
int []rem = { 2, 2 };
int N = div.Length;
Console.WriteLine(findNum(div, rem, N));
}
}
// This code is contributed by ajit
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to implement
// above approach
// Function to find the number
function findNum($div, $rem, $N)
{
$num = $rem[$N - 1];
for ($i = $N - 2; $i >= 0; $i--)
{
$num = $num * $div[$i] + $rem[$i];
}
return $num;
}
// Driver Code
$div = array( 8, 3 );
$rem = array(2, 2 );
$N = sizeof($div);
echo findNum($div, $rem, $N);
// This code is contributed by ajit
?>
java 描述语言
<script>
// Javascript program to implement above approach
// Function to find the number
function findNum(div, rem, N)
{
var num = rem[N - 1];
for (var i = N - 2; i >= 0; i--) {
num = num * div[i] + rem[i];
}
return num;
}
// Driver Code
var div = [ 8, 3 ];
var rem = [ 2, 2 ];
var N = div.length;
document.write( findNum(div, rem, N));
</script>
Output:
18
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