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将矩阵逐步化简至行最简形并求秩
将矩阵逐步化简至行最简形并求秩
支持分数，显示步骤，9x9以内矩阵。
【注意】：由于实时录入字符，请务必用【英文输入法】，且用空格代替回车，【输矩阵时不要按回车（输行列数时可以）】，输错请重新运行程序。
截图:
 
/************************
@author:citrus
@date:2021/1/11
*************************/
#include
#include
#include
#include
#define N 9
int isPrime(int x);
int m=-1,n=-1;
typedef struct fraction
{
int numerator;
int denominator;
}Fraction;
Fraction reduce(Fraction f)
{
if(f.numerator==0)
{
f.denominator=1;
return f;
}
else if(f.denominator!=0)
{
int sign=f.numerator/abs(f.numerator);
f.numerator=abs(f.numerator);
int gcd=1;
for(int i=2;i
Fraction f3={f1.numerator*f2.denominator+f1.denominator*f2.numerator,f1.denominator*f2.denominator};
return reduce(f3);
}
Fraction minus(Fraction f1,Fraction f2)
{
Fraction f3={f1.numerator*f2.denominator-f1.denominator*f2.numerator,f1.denominator*f2.denominator};
return reduce(f3);
}
Fraction time(Fraction f1,Fraction f2)
{
Fraction f3={f1.numerator*f2.numerator,f1.denominator*f2.denominator};
return reduce(f3);
}
Fraction divide(Fraction f1,Fraction f2)
{
int sign=f2.numerator/abs(f2.numerator);
Fraction f3={f1.numerator*f2.denominator*sign,f1.denominator*abs(f2.numerator)};
return reduce(f3);
}
Fraction reciprocal(Fraction f)
{
if(f.numerator!=0&&f.denominator!=0)
{
int sign=f.numerator/abs(f.numerator);
Fraction f0={f.denominator*sign,abs(f.numerator)};
return reduce(f0);
}
}
Fraction absolution(Fraction f)
{
Fraction f0=f;
f0.numerator=abs(f0.numerator);
return reduce(f0);
}
int isPrime(int x)
{
if(x>=-1&&x
for(int i=0;i
mar[i][j].numerator=0;
mar[i][j].denominator=1;
}
}
}
void printMartrix(Fraction mar[][N],char *str)
{
printf("%s ==================================================\n",str);
for(int i=0;i
if(mar[i][j].denominator==1)printf("%4d ",mar[i][j].numerator);
else printf("%3d/%-2d ",mar[i][j].numerator,mar[i][j].denominator);
}
printf("\n");
}
puts("");
}
void printFullMartrix(Fraction mar[][N],char *str)
{
printf("%s ==================================================\n",str);
for(int i=0;i
printf("%3d/%-2d ",mar[i][j].numerator,mar[i][j].denominator);
}
printf("\n");
}
puts("");
}
void inputMartrix(Fraction mar[][N])
{
printf("请输入%d×%d矩阵(支持分数,仅以一空格隔开,不用回车,[实时录入],自动结束,请勿输错,请先切换至EngIME,如-12/5 7 2/3 ):\n",m,n);
int t=0;
int sign=1;
int tofm=0;
int p=1;
int shouldentercount;
int entercount=0;
while(p
puts("");
entercount++;
}
char c=getche();
if(c=='-')
{
sign=-1;
}
else if(c>='0'&&c
mar[(p-1)/n][(p-1)%n].numerator=sign*t;
t=0;
sign=1;
tofm=1;
}
else if(c==' ')
{
if(tofm==1)
{
mar[(p-1)/n][(p-1)%n].denominator=sign*t;
}
else
{
mar[(p-1)/n][(p-1)%n].numerator=sign*t;
}
t=0;
sign=1;
tofm=0;
p++;
}
else if(c==-95||c==-94)
{
puts("\n错误,请结束并切换至EngIME.");
}
}
}
void ktimesRowi(Fraction mar[][N],int k,int i)
{
if(i>=1&&ik,1};
for(int x=0;x
if(i>=1&&i
mar[i][x]=time(mar[i][x],k);
}
}
}
void simplifyRowi(Fraction mar[][N],int i)
{
if(i>=1&&i
if(mar[i][x].numerator!=0)
{
if(gotNZFE==0)
{
gotNZFE=1;
reciprocalOfNonZeroFirstElem=reciprocal(mar[i][x]);
}
if(gotNZFE==1)
{
if(!(reciprocalOfNonZeroFirstElem.numerator==1&&reciprocalOfNonZeroFirstElem.denominator==1))
{
if(printed==0)
{
if(reciprocalOfNonZeroFirstElem.denominator==1&&reciprocalOfNonZeroFirstElem.numerator>0)
printf("r%d × %d",i+1,reciprocalOfNonZeroFirstElem.numerator);
else if(reciprocalOfNonZeroFirstElem.denominator==1&&reciprocalOfNonZeroFirstElem.numerator0)
printf("r%d × %d/%d",i+1,reciprocalOfNonZeroFirstElem.numerator,reciprocalOfNonZeroFirstElem.denominator);
else if(reciprocalOfNonZeroFirstElem.denominator!=1&&reciprocalOfNonZeroFirstElem.numerator
printf("r%d - r%d",i,j);
i--;j--;
for(int x=0;x
if(i>=1&&i=1&&j
mar[i][x]=minus(mar[i][x],time(f,mar[j][x]));
}
printMartrix(mar,"");
}
}
void swapRowiRowj(Fraction mar[][N],int i,int j)
{
if(i>=1&&i=1&&j r%d",i,j);
i--;j--;
Fraction temp;
for(int x=0;x
if(r>=1&&r=1&&c
for(int j=i+1;j
swapRowiRowj(mar,i+1,j+1);
}
}
}
}
}
int haveNonZeroElemFromRowpInColumnq(Fraction mar[][N],int p,int q)
{
if(p>=1&&p=1&&q
if(mar[x][q].numerator!=0)
{
havenonzero=1;
}
}
return havenonzero;
}
}
void simplifyToSimplestRowForm(Fraction mar[][N])
{
int k=1;
int i=1;
while(i>=1&&i=1&&k
for(int q=i;q
if(mar[q-1][k-1].numerator!=0)RowiminusRowj(mar,q,i);
}
i++;
k++;
}
else if(haveNonZeroElemFromRowpInColumnq(mar,i,k)==0)k++;
}
k=1;
i=1;
while(i>=1&&i=1&&k
for(int q=1;q
RowiminusFractionkRowj(mar,q,i,divide(mar[q-1][k-1],mar[i-1][k-1]));
}
}
i++;
k++;
}
else if(haveNonZeroElemFromRowpInColumnq(mar,i,k)==0)k++;
}
}
int rank(Fraction mar[][N])
{
int r=m;
for(int i=m-1;i>=0;i--)
{
int fullrowzero=1;
for(int j=0;j
Fraction mar[N][N];
char c;
do
{
initMartrix(mar);
int t=0;
do
{
if(!t)printf("\n请输入m×n矩阵的m n(1 |