Răspuns:
#include <iostream>
using namespace std;
int main()
{
int n,m,k1=0,k2=0,i,v[51],a,c;
cout<<"n: ";
cin>>n;
cout<<"m: ";
cin>>m;
int n2=n;
while(n>0){
n=n/10;
k2++;
}
for(i=1;i<=k2;i++){
int c=n2%10;
n2=n2/10;
v[i]=c;
}
cout<<endl;
for(i=1;i<k2;i++){
for(int j=i+1;j<=k2;j++){
if(v[i]<v[j]){
a=v[i];
v[i]=v[j];
v[j]=a;
}
}
}
int nr=0;
for(i=1;i<=k2-m;i++){
nr=nr*10+v[i];
}
cout<<nr;
return 0;
}
Explicație:
1.am aflat cate cifre are n;
2.am pus fiecare cifra a lui n intr-un vector;
3.am ordonat crescator vectorul;
4.am format un numar nou eliminând din n exact m cifre
