矩阵化为行阶梯型矩阵 和 化为行最简矩阵 【显示出每一步的步骤，有源码】

#include<iostream> #include<cstdio> #include<iomanip> #include<vector> #include<cmath> using namespace std; const int maxn = 10; const int inf = 0x3f3f3f3f; double a[maxn][maxn]; struct st{ int x,y; }; vector<st>first; // 存首非0元的坐标 int n,m;//n为行 m 为列 int count = 0; void puutDouble(double jian){ // 化为分数 if((int)jian == jian) { printf("%0.1lf",jian); return ; } int j; double xx; for( j = 1;j <= 99;j++){ xx = jian*j; if(xx > 0) xx += 0.005; else xx -= 0.005; if(abs( (int) xx - xx ) <= 0.01 ){ break; } } cout<<(int)xx<<"/"<<j; } void putt(){ //输出 for(int i = 1;i <= n;i++){ for(int j = 1;j <=m ;j++){ if(abs(a[i][j]) < 0.001) a[i][j] = 0; if((int)a[i][j]!=a[i][j]){ cout<<setw(2)<<"("; puutDouble(a[i][j]); cout<<")"; } else cout<<setw(6)<<a[i][j]; // printf("%-.1lf ",a[i][j]); } printf("\n"); } } void divi(int y,int x){//化为阶梯 if(y>n||x>m) return; bool f = 0; if(abs(a[y][x]) < 1 &&a[y][x]!=0){ //判断是否要自乘，目前小于1才自乘（大于一自乘还存在问题） double xx = 1 / a[y][x]; if(xx == 0||abs(xx)<=inf) { // cout<<"r"<<d<<"*"<<x<<endl; for(int i = x;i <= m;i++){ a[y][i] *= xx; } puutDouble(xx); cout<<"*r"<<y<<endl; cout<<"第"<<++count<<"次变换"<<endl; putt(); cout<<"------divi自乘线----------"<<endl; }} int i = y+1; for(int i = y+1;i <= n;i++){ double jian = -a[i][x]*1.0 / a[y][x]*1.0 ; if(jian == 0||abs(jian)>=inf) continue; cout<<"r"<<i<<"+("; puutDouble(jian); cout<<")*"<<"r"<<y<<endl; f = 1; for(int j = x;j <= m;j++){ a[i][j] = a[i][j] + a[y][j]*jian; } } if(f){ cout<<"第"<<++count<<"次变换"<<endl; putt(); cout<<"---------divi1-----------"<<endl; } } void divi2(){//化为最简 int cont = 0; for(int z = 0;z < first.size();z++){ bool f = 0; int y = first[z].y; int x = first[z].x; if(a[y][x] == 0||y == 1) continue; for(int i = y-1;i >= 1;i-- ){ double jian = -a[i][x]*1.0 / a[y][x]*1.0 ; if(jian == 0||jian>=inf) continue; cout<<"r"<<i<<"+("; puutDouble(jian); cout<<")*"<<"r"<<y<<endl; f = 1; for(int j = x;j <= m;j++){ a[i][j] = a[i][j] + a[y][j]*jian; } } if(f){ cout<<"第"<<++cont<<"次变换"<<endl; putt(); cout<<"-------divi2-----------"<<endl; } } } int main() { cout<<"输入矩阵的行数和列数："<<endl; cin>>n>>m; cout<<"输入该矩阵"<<endl; for(int i = 1;i <= n;i++){ for(int j = 1;j <= m;j++){ cin>>a[i][j]; } } cout<<"------------------------"<<endl; for(int i = 1,j = 1;i <= n&&j <= m;i++,j++){ while(a[i][j] == 0) {// 如果下一个对角线为0 则往右移动 j++; if(j>m&&i<n){//如果这一行都为0 那么把下一行交换上来，现在这一行放在最下面，重新检查这一行 j = i; for(int z = 1;z <= m ;z++){ swap(a[i][z],a[n][z]); } for(int c = i;c < n - 1;c++) for(int z = 1;z <=m;z++){ swap(a[c][z],a[c+1][z]); } cout<<"第"<<++count<<"次变换"<<endl; cout<<"r"<<i<<"与"<<"r"<<i+1<<"交换"<<endl; putt(); } } divi(i,j); st a; a.y = i;a.x = j; first.push_back(a); } cout<<"已转换为行阶梯型矩阵"<<endl; cout<<"--------------------"<<endl; divi2(); cout<<"已转换为行最简矩阵"<<endl; system("pause"); return 0; }