Răspuns:
#include <iostream>
#include <cstring>
using namespace std;
char fraza1[101], fraza2[101];
short n1, n2, i, vf1[30],vf2[30];
int main()
{
cin.getline(fraza1,101);
n1=strlen(fraza1);
for (i=0; i<n1; ++i)
if (fraza1[i]>='a' and fraza1[i]<='z') ++vf1[(int)fraza1[i]-97];
cin.getline(fraza2,101);
n2=strlen(fraza2);
for (i=0; i<n2; ++i)
if (fraza2[i]>='a' && fraza2[i]<='z') ++vf2[(int)fraza2[i]-97];
short ok=1;
for (i=0; i<30; ++i)
{
if (vf1[i]!=vf2[i]) { ok=0; break; }
}
if (ok) cout << "DA";
else cout << "NU";
}
Explicație:
am completat vectori de frecventă a literelor din fraze şi verificare că au aceleaşi frecvenţe
