给定一个数n(变量数)和val(变量之和)，找出有多少个这样的非负积分解是可能的。
例子 ：
Input : n = 5, val = 1
Output : 5
Explanation:
x1 + x2 + x3 + x4 + x5 = 1
Number of possible solution are :
(0 0 0 0 1), (0 0 0 1 0), (0 0 1 0 0),
(0 1 0 0 0), (1 0 0 0 0)
Total number of possible solutions are 5
Input : n = 5, val = 4
Output : 70
Explanation:
x1 + x2 + x3 + x4 + x5 = 4
Number of possible solution are:
(1 1 1 1 0), (1 0 1 1 1), (0 1 1 1 1),
(2 1 0 0 1), (2 2 0 0 0)........ so on......
Total numbers of possible solutions are 70提问人：微软访谈
1.进行递归函数调用countSolutions(int n，int val)
2.调用此解决方案函数countSolutions(n-1，val-i)直到n = 1且val> = 0，然后返回1。
下面是上述方法的实现：
C++
// CPP program to find the numbers
// of non negative integral solutions
#include
using namespace std;
// return number of non negative
// integral solutions
int countSolutions(int n, int val)
{
// initialize total = 0
int total = 0;
// Base Case if n = 1 and val >= 0
// then it should return 1
if (n == 1 && val >=0)
return 1;
// iterate the loop till equal the val
for (int i = 0; i <= val; i++){
// total solution of equations
// and again call the recursive
// function Solutions(variable,value)
total += countSolutions(n-1, val-i);
}
// return the total no possible solution
return total;
}
// driver code
int main(){
int n = 5;
int val = 20;
cout< Java
// Java program to find the numbers
// of non negative integral solutions
class GFG {
// return number of non negative
// integral solutions
static int countSolutions(int n, int val) {
// initialize total = 0
int total = 0;
// Base Case if n = 1 and val >= 0
// then it should return 1
if (n == 1 && val >= 0)
return 1;
// iterate the loop till equal the val
for (int i = 0; i <= val; i++) {
// total solution of equations
// and again call the recursive
// function Solutions(variable, value)
total += countSolutions(n - 1, val - i);
}
// return the total no possible solution
return total;
}
// Driver code
public static void main(String[] args) {
int n = 5;
int val = 20;
System.out.print(countSolutions(n, val));
}
}
// This code is contributed by Anant Agarwal.Python3
# Python3 program to find the numbers
# of non negative integral solutions
# return number of non negative
# integral solutions
def countSolutions(n, val):
# initialize total = 0
total = 0
# Base Case if n = 1 and val >= 0
# then it should return 1
if n == 1 and val >=0:
return 1
# iterate the loop till equal the val
for i in range(val+1):
# total solution of equations
# and again call the recursive
# function Solutions(variable,value)
total += countSolutions(n-1, val-i)
# return the total no possible solution
return total
# driver code
n = 5
val = 20
print(countSolutions(n, val))C#
// C# program to find the numbers
// of non negative integral solutions
using System;
class GFG {
// return number of non negative
// integral solutions
static int countSolutions(int n, int val) {
// initialize total = 0
int total = 0;
// Base Case if n = 1 and val >= 0
// then it should return 1
if (n == 1 && val >= 0)
return 1;
// iterate the loop till equal the val
for (int i = 0; i <= val; i++) {
// total solution of equations
// and again call the recursive
// function Solutions(variable, value)
total += countSolutions(n - 1, val - i);
}
// return the total no possible solution
return total;
}
// Driver code
public static void Main()
{
int n = 5;
int val = 20;
Console.WriteLine(countSolutions(n, val));
}
}
// This code is contributed by Anant vt_m.PHP
= 0 then it should
// return 1
if ($n == 1 && $val >=0)
return 1;
// iterate the loop
// till equal the val
for ($i = 0; $i <= $val; $i++)
{
// total solution of equations
// and again call the recursive
// function Solutions(variable,value)
$total += countSolutions($n - 1,
$val - $i);
}
// return the total
// no possible solution
return $total;
}
// Driver Code
$n = 5;
$val = 20;
echo countSolutions($n, $val);
// This code is contributed by nitin mittal.
?>Javascript
输出 ：
10626