方程 x + y + z 的解的数量< = n
给定四个数字 X，Y，z，n，任务是求方程 x + y + z <= n 的解的个数，使得 0 <= x <= X，0 <= y <= Y，0 <= z <= Z。
示例:
Input: x = 1, y = 1, z = 1, n = 1
Output: 4
Input: x = 1, y = 2, z = 3, n = 4
Output: 20
方法:让我们显式迭代 x 和 y 的所有可能值(使用嵌套循环)。对于这样一个 x 和 y 的固定值，问题简化为有多少个 z 值使得z<= n–x–y 和 0 < = z < = Z.
以下是找到解决方案数量所需的实施:
C++
// CPP program to find the number of solutions for
// the equation x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
#include <bits/stdc++.h>
using namespace std;
// function to find the number of solutions for
// the equation x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
int NumberOfSolutions(int x, int y, int z, int n)
{
// to store answer
int ans = 0;
// for values of x
for (int i = 0; i <= x; i++) {
// for values of y
for (int j = 0; j <= y; j++) {
// maximum possible value of z
int temp = n - i - j;
// if z value greater than equals to 0
// then only it is valid
if (temp >= 0) {
// find minimum of temp and z
temp = min(temp, z);
ans += temp + 1;
}
}
}
// return required answer
return ans;
}
// Driver code
int main()
{
int x = 1, y = 2, z = 3, n = 4;
cout << NumberOfSolutions(x, y, z, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find the number of solutions for
// the equation x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
import java.io.*;
class GFG {
// function to find the number of solutions for
// the equation x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
static int NumberOfSolutions(int x, int y, int z, int n)
{
// to store answer
int ans = 0;
// for values of x
for (int i = 0; i <= x; i++) {
// for values of y
for (int j = 0; j <= y; j++) {
// maximum possible value of z
int temp = n - i - j;
// if z value greater than equals to 0
// then only it is valid
if (temp >= 0) {
// find minimum of temp and z
temp = Math.min(temp, z);
ans += temp + 1;
}
}
}
// return required answer
return ans;
}
// Driver code
public static void main (String[] args) {
int x = 1, y = 2, z = 3, n = 4;
System.out.println( NumberOfSolutions(x, y, z, n));
}
}
// this code is contributed by anuj_67..
Python 3
# Python3 program to find the number
# of solutions for the equation
# x + y + z <= n, such that
# 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
# function to find the number of solutions
# for the equation x + y + z <= n, such that
# 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
def NumberOfSolutions(x, y, z, n) :
# to store answer
ans = 0
# for values of x
for i in range(x + 1) :
# for values of y
for j in range(y + 1) :
# maximum possible value of z
temp = n - i - j
# if z value greater than equals
# to 0 then only it is valid
if temp >= 0 :
# find minimum of temp and z
temp = min(temp, z)
ans += temp + 1
# return required answer
return ans
# Driver code
if __name__ == "__main__" :
x, y, z, n = 1, 2, 3, 4
# function calling
print(NumberOfSolutions(x, y, z, n))
# This code is contributed by ANKITRAI1
C
// C# program to find the number of solutions for
// the equation x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
using System;
public class GFG{
// function to find the number of solutions for
// the equation x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
static int NumberOfSolutions(int x, int y, int z, int n)
{
// to store answer
int ans = 0;
// for values of x
for (int i = 0; i <= x; i++) {
// for values of y
for (int j = 0; j <= y; j++) {
// maximum possible value of z
int temp = n - i - j;
// if z value greater than equals to 0
// then only it is valid
if (temp >= 0) {
// find minimum of temp and z
temp = Math.Min(temp, z);
ans += temp + 1;
}
}
}
// return required answer
return ans;
}
// Driver code
static public void Main (){
int x = 1, y = 2, z = 3, n = 4;
Console.WriteLine( NumberOfSolutions(x, y, z, n));
}
}
// This code is contributed by anuj_67..
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find the number
// of solutions for the equation
// x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
// function to find the number of
// solutions for the equation
// x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
function NumberOfSolutions($x, $y, $z, $n)
{
// to store answer
$ans = 0;
// for values of x
for ($i = 0; $i <= $x; $i++)
{
// for values of y
for ($j = 0; $j <= $y; $j++)
{
// maximum possible value of z
$temp = $n - $i - $j;
// if z value greater than equals
// to 0 then only it is valid
if ($temp >= 0)
{
// find minimum of temp and z
$temp = min($temp, $z);
$ans += $temp + 1;
}
}
}
// return required answer
return $ans;
}
// Driver code
$x = 1; $y = 2;
$z = 3; $n = 4;
echo NumberOfSolutions($x, $y, $z, $n);
// This code is contributed
// by Akanksha Rai(Abby_akku)
?>
java 描述语言
<script>
// Javascript program to find the number
// of solutions for the equation x + y + z <= n,
// such that 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
// Function to find the number of solutions for
// the equation x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
function NumberOfSolutions(x, y, z, n)
{
// To store answer
var ans = 0;
// for values of x
for(var i = 0; i <= x; i++)
{
// for values of y
for(var j = 0; j <= y; j++)
{
// Maximum possible value of z
var temp = n - i - j;
// If z value greater than equals to 0
// then only it is valid
if (temp >= 0)
{
// Find minimum of temp and z
temp = Math.min(temp, z);
ans += temp + 1;
}
}
}
// Return required answer
return ans;
}
// Driver Code
var x = 1, y = 2, z = 3, n = 4;
document.write(NumberOfSolutions(x, y, z, n));
// This code is contributed by Ankita saini
</script>
Output:
20
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