P2089 烤鸡

题目背景

猪猪hanke得到了一只鸡

题目描述

猪猪Hanke特别喜欢吃烤鸡（本是同畜牲，相煎何太急！）Hanke吃鸡很特别，为什么特别呢？因为他有10种配料（芥末、孜然等），每种配料可以放1—3克，任意烤鸡的美味程度为所有配料质量之和

现在，Hanke想要知道，如果给你一个美味程度，请输出这10种配料的所有搭配方案

输入输出格式

输入格式：

一行,n<=5000

输出格式：

第一行，方案总数

第二行至结束，10个数，表示每种配料所放的质量

按字典序排列。

如果没有符合要求的方法，就只要在第一行输出一个“0”

#include<iostream>
using namespace std;
int main()
{
    int a, b, c, d, e, f, g, h, i, j, in, x = 0;
    cin >> in;
    for (a = 1; a <= 3; a++)
    {
        for (b = 1; b <= 3; b++)
        {
            for (c = 1; c <= 3; c++)
            {
                for (d = 1; d <= 3; d++)
                {
                    for (e = 1; e <= 3; e++)
                    {
                        for (f = 1; f <= 3; f++)
                        {
                            for (g = 1; g <= 3; g++)
                            {
                                for (h = 1; h <= 3; h++)
                                {
                                    for (i = 1; i <= 3; i++)
                                    {
                                        for (j = 1; j <= 3; j++)
                                        {
                                            if (a + b + c + d + e + f + g + h + i + j == in)
                                            {
                                                x++;
                                            }
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }
    cout << x << endl;
    for (a = 1; a <= 3; a++)
    {
        for (b = 1; b <= 3; b++)
        {
            for (c = 1; c <= 3; c++)
            {
                for (d = 1; d <= 3; d++)
                {
                    for (e = 1; e <= 3; e++)
                    {
                        for (f = 1; f <= 3; f++)
                        {
                            for (g = 1; g <= 3; g++)
                            {
                                for (h = 1; h <= 3; h++)
                                {
                                    for (i = 1; i <= 3; i++)
                                    {
                                        for (j = 1; j <= 3; j++)
                                        {
                                            if (a + b + c + d + e + f + g + h + i + j == in)
                                            {
                                                cout << a << " ";
                                                cout << b << " ";
                                                cout << c << " ";
                                                cout << d << " ";
                                                cout << e << " ";
                                                cout << f << " ";
                                                cout << g << " ";
                                                cout << h << " ";
                                                cout << i << " ";
                                                cout << j << endl;
                                            }
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }
}

原文地址：https://www.cnblogs.com/hsjj/p/P2089.html