找出两个数的适当分数中的第 n 个数字
给定三个整数 P 、 Q 和 N ，其中 P < Q ，任务是计算 P / Q 的分数值，并找到小数点后的 N th 位。 例
输入: P = 1，Q = 2，N = 1 输出: 5 (1 / 2) = 0.5，5 为小数点后第一位数字。 输入: P = 5，Q = 6，N = 5 输出:3 (5/6)= 0.833333333……
方法:初始化一个整数变量 res ，它存储结果的第 n 个数字。现在，当 N > 0 执行以下操作:
- 将 N 减少 1 。
- 要计算数字，更新数值 P = P * 10 。
- 计算 res = P / Q 并更新 P = P % Q 。
最后，打印 res. 下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the Nth digit
// in the fraction (p / q)
int findNthDigit(int p, int q, int N)
{
// To store the resultant digit
int res;
// While N > 0 compute the Nth digit
// by dividing p and q and store the
// result into variable res
// and go to next digit
while (N > 0) {
N--;
p *= 10;
res = p / q;
p %= q;
}
return res;
}
// Driver code
int main()
{
int p = 1, q = 2, N = 1;
cout << findNthDigit(p, q, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to print the Nth digit
// in the fraction (p / q)
static int findNthDigit(int p,
int q, int N)
{
// To store the resultant digit
int res = 0;
// While N > 0 compute the Nth digit
// by dividing p and q and store the
// result into variable res
// and go to next digit
while (N > 0)
{
N--;
p *= 10;
res = p / q;
p %= q;
}
return res;
}
// Driver code
public static void main(String args[])
{
int p = 1, q = 2, N = 1;
System.out.println(findNthDigit(p, q, N));
}
}
// This code is contributed by AnkitRai01
Python 3
# Python3 implementation of the approach
# Function to print the Nth digit
# in the fraction (p / q)
def findNthDigit(p, q, N) :
# While N > 0 compute the Nth digit
# by dividing p and q and store the
# result into variable res
# and go to next digit
while (N > 0) :
N -= 1;
p *= 10;
res = p // q;
p %= q;
return res;
# Driver code
if __name__ == "__main__" :
p = 1; q = 2; N = 1;
print(findNthDigit(p, q, N));
# This code is contributed by kanugargng
C
// C# implementation of the approach
using System;
class GFG
{
// Function to print the Nth digit
// in the fraction (p / q)
static int findNthDigit(int p, int q, int N)
{
// To store the resultant digit
int res = 0;
// While N > 0 compute the Nth digit
// by dividing p and q and store the
// result into variable res
// and go to next digit
while (N > 0)
{
N--;
p *= 10;
res = p / q;
p %= q;
}
return res;
}
// Driver code
public static void Main()
{
int p = 1, q = 2, N = 1;
Console.WriteLine(findNthDigit(p, q, N));
}
}
// This code is contributed by AnkitRai01
java 描述语言
<script>
// JavaScript implementation of the approach
// Function to print the Nth digit
// in the fraction (p / q)
function findNthDigit(p, q, N)
{
// To store the resultant digit
var res;
// While N > 0 compute the Nth digit
// by dividing p and q and store the
// result into variable res
// and go to next digit
while (N > 0)
{
N--;
p *= 10;
res = parseInt(p / q);
p %= q;
}
return res;
}
// Driver code
var p = 1,
q = 2,
N = 1;
document.write(findNthDigit(p, q, N));
// This code is contributed by rdtank.
</script>
Output
5
时间复杂度: O(N)
辅助空间 : O(1)
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