P2089 烤鸡
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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;
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