本文迁移自老博客，原始链接为 https://seven.blog.ustc.edu.cn/usacofeed-ratios/
给出目标饲料 3：4：5 和三种饲料的比例：
1:2:3
3:7:1
2:1:2
编程找出使这三种饲料用量最少的能配出目标饲料的方案，要是不能用这三种饲料调配目标饲料，输出“NONE”。例：
8(1:2:3) + 1(3:7:1) + 5(2:1:2) = (21:28:35) = 7(3:4:5)
解法：由于各种饲料的份数都小于100，所以可以直接暴力枚举,判定时能用乘法，尽量不要用除法。
#include<iostream>
#include<algorithm>
#include<fstream>
using namespace std;
ofstream fout ("ratios.out");
ifstream fin ("ratios.in");
int a , b , c;
int x1 , y1 , z1;
int x2 , y2 , z2;
int x3 , y3 , z3;
int main()
{
fin>>a>>b>>c;
fin>>x1>>y1>>z1;
fin>>x2>>y2>>z2;
fin>>x3>>y3>>z3;
bool flag = false;
for(int i = 0 ; i < 100 && !flag; i++)
{
for(int j = 0 ; j < 100 && !flag;j++)
{
for(int k = 0 ; k < 100 && !flag;k++)
{
int t1 = x1 * i + x2 *j + x3 * k;
int t2 = y1 * i + y2 *j + y3 * k;
int t3 = z1 * i + z2 *j + z3 * k;
int r;
if(a && t1 && t1 % a == 0)
{
r = t1 / a;
if(r * b == t2 && r * c == t3)
{
flag = true;
fout<<i<<" "<<j<<" "<<k<<" "<<r<<endl;
break;
}
}
else if(b && t2 && t2 % b == 0)
{
r = t2 / b;
if(r * a == t1 && r * c == t3)
{
flag = true;
fout<<i<<" "<<j<<" "<<k<<" "<<r<<endl;
break;
}
}
else if(c && t3 && t3 % c == 0)
{
r = t3 / c;
if(r * b == t2 && r * a == t1)
{
flag = true;
fout<<i<<" "<<j<<" "<<k<<" "<<r<<endl;
break;
}
}
}
}
}
if(!flag)
fout<<"NONE"<<endl;
return 0;
}
