Ok, now i'll try to make you understanding my situation. Now i paste you my code :
# include <fstream>
# include <cstring>
# include <cmath>
using namespace std;
#define MAX 105
int N[MAX], i[MAX] ;
int j;
char c[MAX];
ifstream f("free.in");
ofstream g("free.out");
void atrh (int A[], int B[]) // A <- B
{
for (int i = 0; i <= B[0]; ++i) A[i] = B[i];
}
void add (int A[], int B[]) // A <- A + B
{
int i, t = 0;
for (i = 1; i <= A[0] || i <= B[0] || t; i++, t /= 10)
A[i] = ( t += ( i <= A[0] ? A[i] : A[i] = 0 ) + ( i <= B[0] ? B[i] : B[i] = 0 ) ) % 10;
A[0] = i - 1;
}
void sub (int A[], int B[]) // A <- A - B, A >= B
{
int i, t = 0;
for (i = 1; i <= A[0]; i++)
A[i] += ( t = ( A[i] -= ( i <= B[0] ? B[i] : B[i] = 0 ) + t ) < 0 ) * 10;
for (; A[0] > 1 && !A[A[0]]; A[0]--);
}
inline int comp (int A[], int B[]) // A ? B , ? == <, >, =
{
while (A[0] && !A[A[0]]) A[0]--;
while (B[0] && !B[B[0]]) B[0]--;
if (A[0] < B[0]) return -1;
else if (A[0] > B[0]) return 1;
for (int i = A[0]; i > 0; --i)
if (A[i] < B[i]) return -1;
else if (A[i] > B[i]) return 1;
return 0;
}
void atr (int A[], long long X) // A[] <- X
{
for ( A[0] = 0; X ; X /= 10)
A[++A[0]] = X % 10;
}
void atr0 ( int A[] )
{
A[0] = 0;
}
void mulmare (int A[], int B[]) // A <- A * B
{
int i, j, t, C[MAX]; // C <- A * B
memset(C, 0, sizeof(C));
for (i = 1; i <= A[0]; i++)
{
for (t = 0, j = 1; j <= B[0] || t; j++, t /= 10)
C[i + j - 1] = ( t += C[i + j - 1] + A[i] * B[j] ) % 10;
if ( i + j - 2 > C[0] ) C[0] = i + j - 2;
}
memcpy(A, C, sizeof(C)); // A <- C
}
void Shr (int A[], int Count) // A <- A / ( 10 * Count )
{
memmove ( &A[1], &A[Count + 1], sizeof(int) * ( A[0] - Count ) );
A[0] -= Count;
}
void Shl (int A[], int Count) // A <- A * ( 10 * Count )
{
memmove ( &A[Count + 1], &A[1], sizeof(int) * A[0] );
memset ( &A[1], 0, sizeof(int) * Count );
A[0] += Count;
}
void imp (int A[], int B) // A[] <- A[] / B
{
int i, t = 0;
for (i = A[0]; i > 0; i--, t %= B)
A[i] = (t = t * 10 + A[i]) / B;
for (; A[0] > 1 && !A[A[0]]; A[0]--);
}
void impmare (int A[], int B[], int C[]) // C <- A * B rest R
{
int R[MAX];
R[0] = 0, C[0] = A[0];
for (int i = A[0]; i ; i--)
{
Shl (R, 1), R[1] = A[i];
for (C[i] = 0; comp(B,R) != 1 ; ++C[i], sub(R, B)) ;
}
for (; !C[C[0]] && C[0] > 1 ; C[0]--);
}
void mul (int A[], int B) // A[] <- A[] * B
{
int i, t = 0;
for (i = 1; i <= A[0] || t; i++, t /= 10)
A[i] = (t += ( i <= A[0] ? A[i] : A[i] = 0 ) * B) % 10;
A[0] = i - 1;
}
void citire(int x[],char a[])
{
x[0]=strlen(a);
for ( int i = x[0] -1; i >= 0; --i )
x[i+1]=a[x[0]-i-1] - '0';
}
void write( int A[] )
{
if ( A[0] == 0 ) g << "0";
else for (int i = A[0]; i ; --i) g << A[i];
g << "\n";
}
void radical_2 ( int A[], int B[] )
{
int AUX = 0 , i;
int C[MAX], D[MAX], E[MAX], F[MAX];
if ( A[0] & 1 ) AUX = A[A[0]], i = A[0] - 1; // formez prima pereche
else AUX = A[A[0]] * 10 + A[A[0] - 1], i = A[0] - 2;
int aux = (int) sqrt(AUX) ; // aflu nr. cel mai apropiat de prima pereche
atr (B, aux); // atribui rezultatului nr. aflat anterior
if ( A[0] == 1 || A[0] == 2) return ; // daca nr. are 1, 2 cifre, opresc
AUX -= aux * aux; // fac scaderea
atr(E, AUX), Shl(E, 1), atr(F, A[i]), add(E, F); // E = AUX, E *= 10, E += A[i];
// adica formez noul numar, adaugand urm. pereche de cate 2 cifre
Shl(E, 1), atr(F, A[i - 1]), add(E, F); // E *= 10, E += A[i - 1];
// la fel, formez noul numar adaugand cea de-a doua cifra
atrh(C, E), atrh(F, B), mul(F, 2), atrh(E, F); // dublez pe E
i -= 2; // scad 2 unitati, adica cele 2 cifre atribuite
atrh(D, C), Shr(D, 1), impmare(D, E, D); // aici scap de ultima cifra a nr. si o impart la E
int U[] = {1, 9}; // vector auxiliar ce reprezinta cifra 9
int Z[] = {1, 1}; // vector auxiliar ce reprezinta cifra 1
if (comp(D, U) == 1) atr(D, 9); // daca cumva ultima cifra pe care trebuie sa o adaugam la
// rezultat e > 9, atunci ii atribuim valoarea maxima, 9
Shl(E, 1), add(E, D), mulmare(E, D); // E *= 10, E += D, E *= D
// adica adaug la E cifra D si inmultesc nr. format cu D
while ( comp(E, C) == 1 ) // daca E > C, adica ca si numarul format din adaugarea perechilor de cate 2 cifre
sub(D, Z), atrh(F, B), mul(F, 2), atrh(E, F), Shl(E, 1), add(E, D), mulmare(E, D);
// atunci refac numarul, adica scad o unitate la numarul D si refac operatiile
sub(C, E); // fac scaderea, adica numarul format E il scad din C, care a fost numarul format din ad. celor 2 cifre
Shl(B, 1), add(B, D); // B *= 10, B += D , adica adaug la rezultat cifra D
for (i = i; i ; i -= 2) // aici merg cu un for care reprezinta pozitia de unde voi adauga cele 2 cifre
{
atr(E, A[i]), atr0(F), Shl(F, 1), add(F, E); // de aici operatiile se reiau
atr(E, A[i - 1]), Shl(F, 1), add(F, E);
Shl(C, 2), add(C, F), atrh(F, B), mul(F, 2), atrh(E, F);
atrh(D, C), Shr(D, 1), impmare(D, E, D);
if (comp(D, U) == 1) atr(D, 9);
Shl(E, 1), add(E, D), mulmare(E, D);
while ( comp(E, C) == 1 )
sub(D, Z), atrh(F, B), mul(F, 2), atrh(E, F), Shl(E, 1), add(E, D), mulmare(E, D);
sub(C, E);
Shl(B, 1), add(B, D);
}
}
int main()
{
f >> c;
citire(N,c);
radical_2(N,i);
sub(N,i);
write(N);
return 0;
}
I want you ( or anybody who want to help me ) to run this code. But first, create a input called
, in the same directory with the code. ( the code is *.cpp ) Now, in the file newly created introduce next :
( it's just a test ) . Now tell me what's the result in
, if you have .
[LE] And sorry for the code's commentary, it's in romanian.
