Feed Ratios
1998 ACM Finals, Dan Adkins
Farmer John feeds his cows only the finest mixture of cow food, which has three components: Barley, Oats, and Wheat. While he knows the precise mixture of these easily mixable grains, he can not buy that mixture! He buys three other mixtures of the three grains and then combines them to form the perfect mixture.
Given a set of integer ratios barley:oats:wheat, find a way to combine them IN INTEGER MULTIPLES to form a mix with some goal ratio x:y:z.
For example, given the goal 3:4:5 and the ratios of three mixtures:
1:2:3
3:7:1
2:1:2
your program should find some minimum number of integer units (the `mixture') of the first, second, and third mixture that should be mixed together to achieve the goal ratio or print `NONE'. `Minimum number' means the sum of the three non-negative mixture integers is minimized.
For this example, you can combine eight units of mixture 1, one unit of mixture 2, and five units of mixture 3 to get seven units of the goal ratio:
8*(1:2:3) + 1*(3:7:1) + 5*(2:1:2) = (21:28:35) = 7*(3:4:5)
Integers in the goal ratio and mixture ratios are all non-negative and smaller than 100 in magnitude. The number of units of each type of feed in the mixture must be less than 100. The mixture ratios are not linear combinations of each other.
PROGRAM NAME: ratios
INPUT FORMAT
| Line 1: |
Three space separated integers that represent the goal ratios |
| Line 2..4: |
Each contain three space separated integers that represent the ratios of the three mixtures purchased. |
SAMPLE INPUT (file ratios.in)
3 4 5
1 2 3
3 7 1
2 1 2
OUTPUT FORMAT
The output file should contain one line containing four integers or the word `NONE'. The first three integers should represent the number of units of each mixture to use to obtain the goal ratio. The fourth number should be the multiple of the goal ratio obtained by mixing the initial feed using the first three integers as mixing ratios.
SAMPLE OUTPUT (file ratios.out)
8 1 5 7
Analysis
This problem seems to be a deoth search problem, which, as a matter of fact, is truly a mathematical problem. This problem can be represented in forms of matrix multiply or a linear equation set.
Initially, the first line is saved in an array called b[MAX](MAX here is 3, but generally, we can deal with more complicated situations in this way by change the value of MAX).
What the next MAX lines can do is also and may function better with a MAX-level matrix A[MAX][MAX](squred). Firstly, turn the description into equations:
Later, using matrix to translate it:
It is obvious to find the solution of the equation set by Cramer Law. But I nearly forget to tell you another important thing, which is as important as the mathematical method, is that if det(A) is 0 and not all of the elements in b[MAX] are 0, then the answer is NONE. What's more, as a practical problem, it is unbelievable to find the answer which is negative. Both are the edges to determine whether the answer is avaliable.
After this, you may be interested in find det(A), and I will describe it in another post.
Code
/**//*
ID:braytay1
PROG:ratios
LANG:C++
*/
#include <iostream>
#include <cmath>
#include <fstream>
#define MAX 3
#define eps 0.000001
using namespace std;
int det(int a[MAX][MAX])
{
double s=1;
double tmp[MAX][MAX];
for (int i=0;i<MAX;i++){
for (int j=0;j<MAX;j++){
tmp[i][j]=double(a[i][j]);
}
}
for (int k=0;k<MAX-1;k++){
for (int i=k+1;i<MAX;i++){
for (int j=k+1;j<MAX;j++){
tmp[i][j]-=tmp[i][k]*tmp[k][j]/tmp[k][k];
}
}
}
for (int i=0;i<MAX;i++)
s*=tmp[i][i];
int res;
res=int(s);
if (fabs(s-res)<eps) return res;
else {
if (res>0) return res+1;
else return res-1;
}
}
int sp_gcd(int a,int b){
a=abs(a);
b=abs(b);
if (a<b) return a==0?b:sp_gcd(b%a,a);
else return b==0?a:sp_gcd(b,a%b);
}
int gcd(int a[MAX],int s){
int res;
res=sp_gcd(a[0],a[1]);
for (int i=2;i<MAX;i++){
res=sp_gcd(res,a[i]);
}
res=sp_gcd(res,s);
return res;
}
int main(){
ifstream fin("ratios.in");
ofstream fout("ratios.out");
int A[MAX][MAX],b[MAX],x[MAX];
int k,flag_s=0;
for (int i=0;i<MAX;i++){
fin>>b[i];
if (b[i]) flag_s=1;
}
for (int i=0;i<MAX;i++)
for (int j=0;j<MAX;j++) fin>>A[j][i];
k=det(A);
if (k==0&&flag_s) cout<<"NONE"<<endl;
else {
int k_sign;
if (k>0) k_sign=1;
else if (k==0) k_sign=0;
else k_sign=-1;
for (int i=0;i<MAX;i++){
int A_tmp[MAX][MAX];
for (int i1=0;i1<MAX;i1++){
for (int j1=0;j1<MAX;j1++){
if (j1==i) A_tmp[i1][j1]=b[i1];
else A_tmp[i1][j1]=A[i1][j1];
}
}
x[i]=det(A_tmp);
}
int div;
div=gcd(x,k);
for (int i=0;i<MAX;i++){
if (x[i]*k_sign<0) {
fout<<"NONE"<<endl;
return 0;
}
}
for (int i=0;i<MAX;i++){
x[i]=x[i]/div*k_sign;
fout<<x[i]<<" ";
}
k=k/div*k_sign;
fout<<k<<endl;
}
return 0;
}