用给定的不等式方程求“N”个解
找出a1a2a3的值，…a n 满足以下两个条件。
打印 a 1 、a 2 、…、a n 的值，否则为“无解”。
注:可能有一个几个方案，打印其中任意一个。
举例:
Input: n = 5, x = 15, y = 15
Output:
11
1
1
1
1
Input: n = 4, x = 324, y = 77
Output:
74
1
1
1
方法:下面是解决这个问题的分步算法:
- 初始化元素的数量以及 x 和 y 的值。
- 如果 y 小于 n 或者 x 大于 n，则没有1…一个 2 的解。
- 将第一个解打印为 y–n+1，将 1 打印为其余元素的解。
以下是上述方法的实现:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long
// Function to calculate all the solutions
void findsolution(ll n, ll x, ll y)
{
// there is no solutions
if ((y - n + 1) * (y - n + 1) + n - 1 < x || y < n) {
cout << "No solution";
return;
}
// print first element as y-n+1
cout << y - n + 1;
// print rest n-1 elements as 1
while (n-- > 1)
cout << endl
<< 1;
}
// Driver code
int main()
{
// initialize the number of elements
// and the value of x an y
ll n, x, y;
n = 5, x = 15, y = 15;
findsolution(n, x, y);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// java implementation of above approach
import java.io.*;
class GFG {
// Function to calculate all the solutions
static void findsolution(long n, long x, long y)
{
// there is no solutions
if ((y - n + 1) * (y - n + 1) + n - 1 < x || y < n) {
System.out.println( "No solution");
return;
}
// print first element as y-n+1
System.out.println( y - n + 1);
// print rest n-1 elements as 1
while (n-- > 1)
System.out.println( "1");
}
// Driver code
public static void main (String[] args) {
// initialize the number of elements
// and the value of x an y
long n, x, y;
n = 5; x = 15; y = 15;
findsolution(n, x, y);
}
}
// This code is contributed
// by ajit
Python 3
# Python3 implementation of above approach
# Function to calculate all the solutions
def findsolution(n, x, y):
# there is no solutions
if ((y - n + 1) * (y - n + 1) +
n - 1 < x or y < n):
print("No solution");
return;
# print first element as y-n+1
print(y - n + 1);
# print rest n-1 elements as 1
while (n > 1):
print(1);
n -= 1;
# Driver code
# initialize the number of elements
# and the value of x an y
n = 5;
x = 15;
y = 15;
findsolution(n, x, y);
# This code is contributed by mits
C
// C# implementation of above approach
using System;
class GFG
{
// Function to calculate all the solutions
static void findsolution(long n,
long x, long y)
{
// there is no solutions
if ((y - n + 1) * (y - n + 1) +
n - 1 < x || y < n)
{
Console.WriteLine( "No solution");
return;
}
// print first element as y-n+1
Console.WriteLine( y - n + 1);
// print rest n-1 elements as 1
while (n-- > 1)
Console.WriteLine( "1");
}
// Driver code
static public void Main ()
{
// initialize the number of elements
// and the value of x an y
long n, x, y;
n = 5; x = 15; y = 15;
findsolution(n, x, y);
}
}
// This code is contributed
// by ajit
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of above approach
// Function to calculate all the solutions
function findsolution($n, $x, $y)
{
// there is no solutions
if (($y - $n + 1) * ($y - $n + 1) +
$n - 1 < $x || $y < $n)
{
echo "No solution";
return;
}
// print first element as y-n+1
echo $y - $n + 1;
// print rest n-1 elements as 1
while ($n-- > 1)
echo "\n" . 1;
}
// Driver code
// initialize the number of elements
// and the value of x an y
$n = 5; $x = 15; $y = 15;
findsolution($n, $x, $y);
// This code is contributed
// by Akanksha Rai(Abby_akku)
java 描述语言
<script>
// Javascript implementation of above approach
// Function to calculate all the solutions
function findsolution(n, x, y)
{
// there is no solutions
if ((y - n + 1) * (y - n + 1) +
n - 1 < x || y < n)
{
document.write( "No solution");
return;
}
// print first element as y-n+1
document.write( y - n + 1);
// print rest n-1 elements as 1
while (n-- > 1)
document.write( "<br>" + 1);
}
// Driver code
// initialize the number of elements
// and the value of x an y
let n = 5;
let x = 15;
let y = 15;
findsolution(n, x, y);
// This code is contributed
// by bobby
</script>
Output:
11
1
1
1
1
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