【USACO3.2.4】饲料调配 纯整数高斯消元

其实就是高斯消元。 


但是常数项未知，但是常数项一定是题目给定的常数的整数倍（K倍），所以要穷举一下这个K，然后做高斯消元。

但是如何判断NONE呢？省事，多循环几次，一直不出解，就是无解了。

最近得写一个高斯消元的模板了……

/*TASK:ratiosLANG:C++*/#include <iostream>#include <cmath>#include <cstdio>using namespace std;const int max_n = 5;int equ, var;// 方程数量， var个变元  所以增广矩阵有equ行，var +1 列 （最后一列为你懂的）int a[max_n][max_n], ta[max_n][max_n];int x[max_n]; //解集int gcd(int a, int b){return a == 0 ? b : gcd(b % a , a);}int lcm(int a, int b){return a / gcd(a, b) * b;}bool guess(){int row=0, col=0; //行，列for (; row != equ && col != var; ++ row, ++ col){int max_row = row;for (int i = row + 1; i != equ; ++ i)if (a[i][col] > a[max_row][col])max_row = i;if (!a[max_row][col])return false;for (int i = col; i <= var; ++ i)swap(a[row][i], a[max_row][i]);for (int i = row + 1; i != equ; ++ i){if (!a[i][col])continue;int tmp = lcm(abs(a[row][col]), abs(a[i][col]));int ta = tmp / a[row][col]; //正负，为他们自己的符号int tb = tmp / a[i][col];for (int j = col; j <= var; ++ j)a[i][j] = a[i][j] * tb - a[row][j] * ta;}}row = equ - 1, col = var - 1;for (;row >= 0; -- row, -- col){int tmp = a[row][var];for (int i = col + 1; i != var; ++ i)tmp -= a[row][i] * x[i];if (tmp % a[row][col])return false;x[col] = tmp / a[row][col];if (x[col] < 0)return false;}return true;}void init(){for (int i = 0; i != 3; ++ i)cin >> ta[i][var];for (int i = 0; i != 3; ++ i)for (int j = 0; j != 3; ++ j) cin >> ta[j][i];}void doit(){for (int ans = 1; ans <= 100; ++ ans){for (int i = 0; i != 3; ++ i)for (int j = 0; j != 4; ++ j)a[i][j] = ta[i][j];for (int i = 0; i != 3; ++ i)a[i][3] *= ans;if (guess()){for (int i = 0; i != var; ++ i)cout<<x[i]<<" ";cout<<ans<<endl;return;}}cout<<"NONE"<<endl;}int main(){//freopen("ratios.in","r",stdin);//freopen("ratios.out","w",stdout);equ = var = 3;init();doit();return 0;}