c++实现线性代数矩阵行简化

输入一个矩阵，可分别输出该矩阵的阶梯型和最简型。

输入仅支持整数，支持分数形式输出。

妈妈再也不用担心俺的线性代数作业～

使用实例：

(实现格式化输出部分写的极为丑陋......)

#include<iostream>
#include<cmath>
#include<algorithm>
#include<iomanip>
#include<cstdio>
#include<string>
#define f(i,l,r) for(i=(l);i<=(r);i++)
#define ff(i,r,l) for(i=(r);i>=(l);i--)
#define ll long long
#define EPS 1e-6
using namespace std;
const int MAXN=105;
int n,m;
char output[MAXN];
string s;
struct frac{
    int x,y=1;
    bool operator < (const frac &tmp)const{
        return 1.0*x/y<1.0*tmp.x/tmp.y;
    }
    frac operator - (const frac &tmp){
        frac ans;
        ans.x=x*tmp.y-y*tmp.x;
        ans.y=y*tmp.y;
        ans.sim();
        return ans;
    }
    frac operator * (const frac &tmp){
        frac ans;
        ans.x=x*tmp.x;
        ans.y=y*tmp.y;
        ans.sim();
        return ans;
    }
    frac operator / (const frac &tmp){
        frac ans;
        ans.x=x*tmp.y;
        ans.y=y*tmp.x;
        ans.sim();
        return ans;
    }
    int gcd(int a,int b){
        return b==0?a:gcd(b,a%b);
    }
    void sim(){
        int d=gcd(x,y);
        x/=d;
        y/=d;
        if(x<0&&y<0){
            x=-x;
            y=-y;
        }
    }
    void write()
    {
        int i;
        char tmp[MAXN];
        int num=0;
        int nx=x,ny=y;
        s="";
        if(!nx){
            s+="0";
        }
        else{
            if(nx<0||ny<0){
                s+='-';
                nx=abs(nx);
                ny=abs(ny);
            }
            if(nx%ny==0){
                nx/=ny;
                while(nx){
                    tmp[++num]='0'+nx%10;
                    nx/=10;
                }
                ff(i,num,1){
                    s+=tmp[i];
                }
            }
            else{
                while(nx){
                    tmp[++num]='0'+nx%10;
                    nx/=10;
                }
                ff(i,num,1){
                    s+=tmp[i];
                }
                s+='/';
                num=0;
                while(ny){
                    tmp[++num]='0'+ny%10;
                    ny/=10;
                }
                ff(i,num,1){
                    s+=tmp[i];
                }
            }
        }
        f(i,s.length(),7){
            cout<<" ";
        }
        cout<<s;
    }
}a[MAXN][MAXN];
void out_f()
{
    int i,j;
    f(i,1,n){
        f(j,1,m){
            cout<<1.0*a[i][j].x/a[i][j].y<<" ";
        }
        cout<<endl;
    }
}
void out()
{
    int i,j;
    f(i,1,n){
        f(j,1,m){
            a[i][j].write();
            cout<<" ";
        }
        cout<<endl;
    }
}
int find(int r,int c)
{
    int i,t=-1;
    f(i,r,n){
        if(t==-1||a[t][c]<a[i][c]) t=i;
    }
    return t;
}
void interchange(int r1,int r2)
{
    int j;
    f(j,1,m){
        swap(a[r1][j],a[r2][j]);
    }
    return;
}
void scale(int r,int c)
{
    int j;
    ff(j,m,c){
        a[r][j]=a[r][j]/a[r][c];
    }
    return;
}
void muilt(int r,int c)
{
    int i,j;
    f(i,r+1,n){
        if(!a[i][c].x) continue;
        ff(j,m,c){
            a[i][j]=a[i][j]-(a[i][c]*a[r][j]);
        }
    }
    return;
}
void gause_1()
{
    int c,r=1;
    f(c,1,n){
        int pos=find(r,c);
        if(!a[pos][c].x) continue;
        interchange(r,pos);
        scale(r,c);
        muilt(r,c);
        r++;
    }
}
void gause_2()
{
    int i,j;
    ff(i,n,1){
        f(j,i+1,n){
            a[i][n+1]=a[i][n+1]-(a[i][j]*a[j][n+1]);
            a[i][j].x=0;
        }
    }
    return;
}
int main()
{
    int i,j;
    cout<<"输入矩阵的行数和列数："<<endl;
    cin>>n>>m;
    cout<<"输入矩阵元素："<<endl;
    f(i,1,n){
        f(j,1,m){
            cin>>a[i][j].x;
        }
    }
    gause_1();
    cout<<"阶梯型为："<<endl;
    out();
    gause_2();
    cout<<"最简型为："<<endl;
    out();
    return 0;
}