用和 N 打印所有可能的连续数字
原文:https://www . geeksforgeeks . org/print-所有可能的连续数字-带 sum-n/
给定一个数字 n。任务是打印所有可能的连续数字，这些数字加起来就是 n
示例:
Input: N = 100
Output:
9 10 11 12 13 14 15 16
18 19 20 21 22
Input: N = 125
Output:
8 9 10 11 12 13 14 15 16 17
23 24 25 26 27
62 63
一个重要的事实是，我们找不到 N/2 以上加起来等于 N 的连续数字，因为 N/2 + (N/2 + 1)将大于 N。因此，我们从 start = 1 开始，直到 end = N/2，并检查每个连续序列是否加起来等于 N。如果是，那么我们打印该序列，并通过增加起点开始寻找下一个序列。
C++
// C++ program to print consecutive sequences
// that add to a given value
#include<bits/stdc++.h>
using namespace std;
void findConsecutive(int N)
{
// Note that we don't ever have to sum
// numbers > ceil(N/2)
int start = 1, end = (N+1)/2;
// Repeat the loop from bottom to half
while (start < end)
{
// Check if there exist any sequence
// from bottom to half which adds up to N
int sum = 0;
for (int i = start; i <= end; i++)
{
sum = sum + i;
// If sum = N, this means consecutive
// sequence exists
if (sum == N)
{
// found consecutive numbers! print them
for (int j = start; j <= i; j++)
cout <<" "<< j;
cout <<"\n";
break;
}
// if sum increases N then it can not exist
// in the consecutive sequence starting from
// bottom
if (sum > N)
break;
}
sum = 0;
start++;
}
}
// Driver code
int main(void)
{
int N = 125;
findConsecutive(N);
return 0;
}
// This code is contributed by shivanisinghss2110
C
// C++ program to print consecutive sequences
// that add to a given value
#include<stdio.h>
void findConsecutive(int N)
{
// Note that we don't ever have to sum
// numbers > ceil(N/2)
int start = 1, end = (N+1)/2;
// Repeat the loop from bottom to half
while (start < end)
{
// Check if there exist any sequence
// from bottom to half which adds up to N
int sum = 0;
for (int i = start; i <= end; i++)
{
sum = sum + i;
// If sum = N, this means consecutive
// sequence exists
if (sum == N)
{
// found consecutive numbers! print them
for (int j = start; j <= i; j++)
printf("%d ", j);
printf("\n");
break;
}
// if sum increases N then it can not exist
// in the consecutive sequence starting from
// bottom
if (sum > N)
break;
}
sum = 0;
start++;
}
}
// Driver code
int main(void)
{
int N = 125;
findConsecutive(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to print
// consecutive sequences
// that add to a given value
import java.io.*;
class GFG
{
static void findConsecutive(int N)
{
// Note that we don't
// ever have to sum
// numbers > ceil(N/2)
int start = 1;
int end = (N + 1) / 2;
// Repeat the loop
// from bottom to half
while (start < end)
{
// Check if there exist
// any sequence from
// bottom to half which
// adds up to N
int sum = 0;
for (int i = start; i <= end; i++)
{
sum = sum + i;
// If sum = N, this means
// consecutive sequence exists
if (sum == N)
{
// found consecutive
// numbers! print them
for (int j = start; j <= i; j++)
System.out.print(j + " ");
System.out.println();
break;
}
// if sum increases N then
// it can not exist in the
// consecutive sequence
// starting from bottom
if (sum > N)
break;
}
sum = 0;
start++;
}
}
// Driver code
public static void main (String[] args)
{
int N = 125;
findConsecutive(N);
}
}
// This code is contributed by m_kit
Python 3
# Python3 program to print consecutive
# sequences that add to a given value
def findConsecutive(N):
# Note that we don't ever have to
# Sum numbers > ceil(N/2)
start = 1
end = (N + 1) // 2
# Repeat the loop from bottom to half
while (start < end):
# Check if there exist any sequence
# from bottom to half which adds up to N
Sum = 0
for i in range(start, end + 1):
Sum = Sum + i
# If Sum = N, this means consecutive
# sequence exists
if (Sum == N):
# found consecutive numbers! prthem
for j in range(start, i + 1):
print(j, end = " ")
print()
break
# if Sum increases N then it can not
# exist in the consecutive sequence
# starting from bottom
if (Sum > N):
break
Sum = 0
start += 1
# Driver code
N = 125
findConsecutive(N)
# This code is contributed by Mohit kumar 29
C
// C# program to print
// consecutive sequences
// that add to a given value
using System;
class GFG
{
static void findConsecutive(int N)
{
// Note that we don't
// ever have to sum
// numbers > ceil(N/2)
int start = 1;
int end = (N + 1) / 2;
// Repeat the loop
// from bottom to half
while (start < end)
{
// Check if there exist
// any sequence from
// bottom to half which
// adds up to N
int sum = 0;
for (int i = start; i <= end; i++)
{
sum = sum + i;
// If sum = N, this means
// consecutive sequence exists
if (sum == N)
{
// found consecutive
// numbers! print them
for (int j = start; j <= i; j++)
Console.Write(j + " ");
Console.WriteLine();
break;
}
// if sum increases N then
// it can not exist in the
// consecutive sequence
// starting from bottom
if (sum > N)
break;
}
sum = 0;
start++;
}
}
// Driver code
static public void Main ()
{
int N = 125;
findConsecutive(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 print consecutive
// sequences that add to a given value
function findConsecutive($N)
{
// Note that we don't ever have
// to sum numbers > ceil(N/2)
$start = 1;
$end = ($N + 1) / 2;
// Repeat the loop from
// bottom to half
while ($start < $end)
{
// Check if there exist any
// sequence from bottom to
// half which adds up to N
$sum = 0;
for ($i = $start; $i <= $end; $i++)
{
$sum = $sum + $i;
// If sum = N, this means
// consecutive sequence exists
if ($sum == $N)
{
// found consecutive
// numbers! print them
for ($j = $start; $j <= $i; $j++)
echo $j ," ";
echo "\n";
break;
}
// if sum increases N then it
// can not exist in the consecutive
// sequence starting from bottom
if ($sum > $N)
break;
}
$sum = 0;
$start++;
}
}
// Driver code
$N = 125;
findConsecutive($N);
// This code is contributed by ajit
?>
java 描述语言
<script>
// Javascript program to print consecutive sequences
// that add to a given value
function findConsecutive( N)
{
// Note that we don't ever have to sum
// numbers > ceil(N/2)
let start = 1, end = Math.trunc((N+1)/2);
// Repeat the loop from bottom to half
while (start < end)
{
// Check if there exist any sequence
// from bottom to half which adds up to N
let sum = 0;
for (let i = start; i <= end; i++)
{
sum = sum + i;
// If sum = N, this means consecutive
// sequence exists
if (sum == N)
{
// found consecutive numbers! print them
for (let j = start; j <= i; j++)
document.write(j+" ");
document.write("<br/>");
break;
}
// if sum increases N then it can not exist
// in the consecutive sequence starting from
// bottom
if (sum > N)
break;
}
sum = 0;
start++;
}
}
// Driver program
let N = 125;
findConsecutive(N);
</script>
输出:
8 9 10 11 12 13 14 15 16 17
23 24 25 26 27
62 63
优化解: 在上面的解中，我们从头到尾一直在重新计算和，这导致了 O(N^2)最坏情况下的时间复杂度。这可以通过使用预先计算的总和数组来避免，或者更好的方法是——只需跟踪到目前为止的总和，并根据它与所需总和的比较情况来调整它。 下面代码的时间复杂度为 O(N)。
C++
// Optimized C++ program to find sequences of all consecutive
// numbers with sum equal to N
#include <bits/stdc++.h>
using namespace std;
void printSums(int N)
{
int start = 1, end = 1;
int sum = 1;
while (start <= N/2)
{
if (sum < N)
{
end += 1;
sum += end;
}
else if (sum > N)
{
sum -= start;
start += 1;
}
else if (sum == N)
{
for (int i = start; i <= end; ++i)
cout <<" "<< i;
cout <<"\n";
sum -= start;
start += 1;
}
}
}
// Driver Code
int main()
{
printSums(125);
return 0;
}
// this code is contributrd by shivanisinghss2110
C
// Optimized C program to find sequences of all consecutive
// numbers with sum equal to N
#include <stdio.h>
void printSums(int N)
{
int start = 1, end = 1;
int sum = 1;
while (start <= N/2)
{
if (sum < N)
{
end += 1;
sum += end;
}
else if (sum > N)
{
sum -= start;
start += 1;
}
else if (sum == N)
{
for (int i = start; i <= end; ++i)
printf("%d ", i);
printf("\n");
sum -= start;
start += 1;
}
}
}
// Driver Code
int main()
{
printSums(125);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Optimized Java program to find
// sequences of all consecutive
// numbers with sum equal to N
import java.io.*;
class GFG {
static void printSums(int N)
{
int start = 1, end = 1;
int sum = 1;
while (start <= N/2)
{
if (sum < N)
{
end += 1;
sum += end;
}
else if (sum > N)
{
sum -= start;
start += 1;
}
else if (sum == N)
{
for (int i = start;
i <= end; ++i)
System.out.print(i
+ " ");
System.out.println();
sum -= start;
start += 1;
}
}
}
// Driver Code
public static void main (String[] args)
{
printSums(125);
}
}
// This code is contributed by anuj_67.
Python 3
# Optimized Python3 program to find sequences of all consecutive
# numbers with sum equal to N
def findConsecutive(N):
start = 1
end = 1
sum = 1
while start <= N/2:
if sum < N:
end += 1
sum += end
if sum > N:
sum -= start
start += 1
if sum == N:
for i in range(start, end + 1):
print(i, end=' ')
print( )
sum -= start
start += 1
# Driver code
N = 125
findConsecutive(N)
# This code is contributed by Sanskriti Agrawal
C
// Optimized C# program to find
// sequences of all consecutive
// numbers with sum equal to N
using System;
class GFG {
static void printSums(int N)
{
int start = 1, end = 1;
int sum = 1;
while (start <= N/2)
{
if (sum < N)
{
end += 1;
sum += end;
}
else if (sum > N)
{
sum -= start;
start += 1;
}
else if (sum == N)
{
for (int i = start;
i <= end; ++i)
Console.Write(i
+ " ");
Console.WriteLine();
sum -= start;
start += 1;
}
}
}
// Driver Code
public static void Main ()
{
printSums(125);
}
}
// This code is contributed by anuj_67.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// Optimized PHP program to find
// sequences of all consecutive
// numbers with sum equal to N
function printSums($N)
{
$start = 1; $end = 1;
$sum = 1;
while ($start <= $N / 2)
{
if ($sum < $N)
{
$end += 1;
$sum += $end;
}
else if ($sum > $N)
{
$sum -= $start;
$start += 1;
}
else if ($sum == $N)
{
for ($i = $start; $i <= $end; ++$i)
echo $i," ";
echo "\n";
$sum -= $start;
$start += 1;
}
}
}
// Driver Code
printSums(125);
// This code is contributed by jit_t
?>
java 描述语言
<script>
// Javascript program to find
// sequences of all consecutive
// numbers with sum equal to N
function printSums(N)
{
let start = 1, end = 1;
let sum = 1;
while (start <= N / 2)
{
if (sum < N)
{
end += 1;
sum += end;
}
else if (sum > N)
{
sum -= start;
start += 1;
}
else if (sum == N)
{
for(let i = start;
i <= end; ++i)
document.write(i + " ");
document.write("<br/>");
sum -= start;
start += 1;
}
}
}
// Driver code
printSums(125);
// This code is contributed by splevel62
</script>
输出:
8 9 10 11 12 13 14 15 16 17
23 24 25 26 27
62 63
参考:T2https://www.careercup.com/page?PID = Microsoft-面试-提问& n=2 本文由nitesh Kumar供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以用write.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 review-team@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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