给定一个正整数X(1≤X≤10 6)中,任务是找到最小值N,使得第一N的自然数的总和为≥X。
例子:
Input: X = 14
Output: 5
Explanation: Sum of first 5 natural numbers is 15 which is greater than X( = 14).
- 1 + 2 = 3( < 14)
- 1 + 2 + 3 = 6( < 14)
- 1 + 2 + 3 + 4 = 10( < 15)
- 1 + 2 + 3 + 4 + 5 = 15( > 14)
Input: X = 91
Output: 13
简单方法:为了解决这个问题的最简单的方法是检查在范围[1,X]的每一个值,并从该范围内的量,第一N的自然数的总和被发现是≥X返回第一个值。
下面是上述方法的实现:
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
// Fucntion to check if sum of first
// N natural numbers is >= X
bool isGreaterEqual(int N, int X)
{
return (N * 1LL * (N + 1) / 2) >= X;
}
// Finds minimum value of
// N such that sum of first
// N natural number >= X
int minimumPossible(int X)
{
for (int i = 1; i <= X; i++) {
// Check if sum of first i
// natural number >= X
if (isGreaterEqual(i, X))
return i;
}
}
// Driver Code
int main()
{
// Input
int X = 14;
// Finds minimum value of
// N such that sum of first
// N natural number >= X
cout << minimumPossible(X);
return 0;
}
Java
// Java Program to implement
// the above approach
import java.io.*;
class GFG
{
// Fucntion to check if sum of first
// N natural numbers is >= X
static boolean isGreaterEqual(int N, int X)
{
return (N * (N + 1) / 2) >= X;
}
// Finds minimum value of
// N such that sum of first
// N natural number >= X
static int minimumPossible(int X)
{
for (int i = 1; i <= X; i++)
{
// Check if sum of first i
// natural number >= X
if (isGreaterEqual(i, X))
return i;
}
return 0;
}
// Driver Code
public static void main (String[] args)
{
// Input
int X = 14;
// Finds minimum value of
// N such that sum of first
// N natural number >= X
System.out.print(minimumPossible(X));
}
}
// This code is contributed by Dharanendra L V.
Python3
# Python3 Program to implement
# the above approach
# Function to check if sum of first
# N natural numbers is >= X
def isGreaterEqual(N, X):
return (N * (N + 1) // 2) >= X
# Finds minimum value of
# N such that sum of first
# N natural number >= X
def minimumPossible(X):
for i in range(1, X + 1):
# Check if sum of first i
# natural number >= X
if (isGreaterEqual(i, X)):
return i
# Driver Code
if __name__ == '__main__':
# Input
X = 14
# Finds minimum value of
# N such that sum of first
# N natural number >= X
print (minimumPossible(X))
# This code is contributed by mohit kumar 29.
C#
// C# Program to implement
// the above approach
using System;
public class GFG
{
// Fucntion to check if sum of first
// N natural numbers is >= X
static bool isGreaterEqual(int N, int X)
{
return (N * (N + 1) / 2) >= X;
}
// Finds minimum value of
// N such that sum of first
// N natural number >= X
static int minimumPossible(int X)
{
for (int i = 1; i <= X; i++)
{
// Check if sum of first i
// natural number >= X
if (isGreaterEqual(i, X))
return i;
}
return 0;
}
// Driver Code
static public void Main ()
{
// Input
int X = 14;
// Finds minimum value of
// N such that sum of first
// N natural number >= X
Console.Write(minimumPossible(X));
}
}
// This code is contributed by Dharanendra L V.
Javascript
C++
#include
using namespace std;
// Function to check if sum of first
// N natural numbers is >= X
bool isGreaterEqual(int N, int X)
{
return (N * 1LL * (N + 1) / 2) >= X;
}
// Finds minimum value of
// N such that sum of first
// N natural number >= X
int minimumPossible(int X)
{
int low = 1, high = X, res = -1;
// Binary Search
while (low <= high) {
int mid = low + (high - low) / 2;
// Checks if sum of first 'mid' natural
// numbers is greater than equal to X
if (isGreaterEqual(mid, X)) {
// Update res
res = mid;
// Update high
high = mid - 1;
}
else
// Update low
low = mid + 1;
}
return res;
}
// Driver Code
int main()
{
// Input
int X = 14;
// Finds minimum value of
// N such that sum of first
// N natural number >= X
cout << minimumPossible(X);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to check if sum of first
// N natural numbers is >= X
static boolean isGreaterEqual(int N, int X)
{
return (N * (N + 1) / 2) >= X;
}
// Finds minimum value of
// N such that sum of first
// N natural number >= X
static int minimumPossible(int X)
{
int low = 1, high = X, res = -1;
// Binary Search
while (low <= high)
{
int mid = low + (high - low) / 2;
// Checks if sum of first 'mid' natural
// numbers is greater than equal to X
if (isGreaterEqual(mid, X))
{
// Update res
res = mid;
// Update high
high = mid - 1;
}
else
// Update low
low = mid + 1;
}
return res;
}
// Driver Code
public static void main(String[] args)
{
// Input
int X = 14;
// Finds minimum value of
// N such that sum of first
// N natural number >= X
System.out.print( minimumPossible(X));
}
}
// This code is contributed by code_hunt.
Python3
# Function to check if sum of first
# N natural numbers is >= X
def isGreaterEqual(N, X):
return (N * (N + 1) // 2) >= X;
# Finds minimum value of
# N such that sum of first
# N natural number >= X
def minimumPossible(X):
low = 1
high = X
res = -1;
# Binary Search
while (low <= high):
mid = low + (high - low) // 2;
# Checks if sum of first 'mid' natural
# numbers is greater than equal to X
if (isGreaterEqual(mid, X)):
# Update res
res = mid;
# Update high
high = mid - 1;
else:
# Update low
low = mid + 1;
return res
# Driver Code
if __name__ == "__main__":
# Input
X = 14;
# Finds minimum value of
# N such that sum of first
# N natural number >= X
print(minimumPossible(X));
# This code is contributed by chitranayal.
C#
// C# program for the above approach
using System;
class GFG{
// Function to check if sum of first
// N natural numbers is >= X
static bool isGreaterEqual(int N, int X)
{
return (N * (N + 1) / 2) >= X;
}
// Finds minimum value of
// N such that sum of first
// N natural number >= X
static int minimumPossible(int X)
{
int low = 1, high = X, res = -1;
// Binary Search
while (low <= high)
{
int mid = low + (high - low) / 2;
// Checks if sum of first 'mid' natural
// numbers is greater than equal to X
if (isGreaterEqual(mid, X))
{
// Update res
res = mid;
// Update high
high = mid - 1;
}
else
// Update low
low = mid + 1;
}
return res;
}
// Driver Code
static public void Main()
{
// Input
int X = 14;
// Finds minimum value of
// N such that sum of first
// N natural number >= X
Console.Write( minimumPossible(X));
}
}
// This code is contributed by susmitakundugoaldanga.
输出:
5
时间复杂度: O(N)
辅助空间: O(1)
高效方法:以下是上述方法的实现:
- 这个想法是使用二进制搜索来解决这个问题。
- 初始化变量low = 1,high = X并在此范围内执行二进制搜索。
- 计算中位数=低+(高-低)/ 2,然后检查第一个中位数的总和是否大于或等于x。
- 如果sum≥X ,则将其存储在变量res中,并设置为high = mid-1
- 否则,设置低=中+ 1
- 打印res,这是必需的答案。
下面是上述方法的实现:
C++
#include
using namespace std;
// Function to check if sum of first
// N natural numbers is >= X
bool isGreaterEqual(int N, int X)
{
return (N * 1LL * (N + 1) / 2) >= X;
}
// Finds minimum value of
// N such that sum of first
// N natural number >= X
int minimumPossible(int X)
{
int low = 1, high = X, res = -1;
// Binary Search
while (low <= high) {
int mid = low + (high - low) / 2;
// Checks if sum of first 'mid' natural
// numbers is greater than equal to X
if (isGreaterEqual(mid, X)) {
// Update res
res = mid;
// Update high
high = mid - 1;
}
else
// Update low
low = mid + 1;
}
return res;
}
// Driver Code
int main()
{
// Input
int X = 14;
// Finds minimum value of
// N such that sum of first
// N natural number >= X
cout << minimumPossible(X);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to check if sum of first
// N natural numbers is >= X
static boolean isGreaterEqual(int N, int X)
{
return (N * (N + 1) / 2) >= X;
}
// Finds minimum value of
// N such that sum of first
// N natural number >= X
static int minimumPossible(int X)
{
int low = 1, high = X, res = -1;
// Binary Search
while (low <= high)
{
int mid = low + (high - low) / 2;
// Checks if sum of first 'mid' natural
// numbers is greater than equal to X
if (isGreaterEqual(mid, X))
{
// Update res
res = mid;
// Update high
high = mid - 1;
}
else
// Update low
low = mid + 1;
}
return res;
}
// Driver Code
public static void main(String[] args)
{
// Input
int X = 14;
// Finds minimum value of
// N such that sum of first
// N natural number >= X
System.out.print( minimumPossible(X));
}
}
// This code is contributed by code_hunt.
Python3
# Function to check if sum of first
# N natural numbers is >= X
def isGreaterEqual(N, X):
return (N * (N + 1) // 2) >= X;
# Finds minimum value of
# N such that sum of first
# N natural number >= X
def minimumPossible(X):
low = 1
high = X
res = -1;
# Binary Search
while (low <= high):
mid = low + (high - low) // 2;
# Checks if sum of first 'mid' natural
# numbers is greater than equal to X
if (isGreaterEqual(mid, X)):
# Update res
res = mid;
# Update high
high = mid - 1;
else:
# Update low
low = mid + 1;
return res
# Driver Code
if __name__ == "__main__":
# Input
X = 14;
# Finds minimum value of
# N such that sum of first
# N natural number >= X
print(minimumPossible(X));
# This code is contributed by chitranayal.
C#
// C# program for the above approach
using System;
class GFG{
// Function to check if sum of first
// N natural numbers is >= X
static bool isGreaterEqual(int N, int X)
{
return (N * (N + 1) / 2) >= X;
}
// Finds minimum value of
// N such that sum of first
// N natural number >= X
static int minimumPossible(int X)
{
int low = 1, high = X, res = -1;
// Binary Search
while (low <= high)
{
int mid = low + (high - low) / 2;
// Checks if sum of first 'mid' natural
// numbers is greater than equal to X
if (isGreaterEqual(mid, X))
{
// Update res
res = mid;
// Update high
high = mid - 1;
}
else
// Update low
low = mid + 1;
}
return res;
}
// Driver Code
static public void Main()
{
// Input
int X = 14;
// Finds minimum value of
// N such that sum of first
// N natural number >= X
Console.Write( minimumPossible(X));
}
}
// This code is contributed by susmitakundugoaldanga.
输出:
5
时间复杂度: O(log(X))
辅助空间: O(1)