### 11371 - Number Theory for Newbies

Posted:

**Sat Dec 29, 2007 8:41 pm**Code: Select all

```
thanks to sohel vai and fpavetic.
```

The Online Judge board

https://uva.onlinejudge.org/board/

https://uva.onlinejudge.org/board/viewtopic.php?f=41&t=26752

Page **1** of **5**

Posted: **Sat Dec 29, 2007 8:41 pm**

Code: Select all

```
thanks to sohel vai and fpavetic.
```

Posted: **Sat Dec 29, 2007 8:50 pm**

I don't think you are handling the 'leading 0 case' properly.

What is your output for 12300?

What is your output for 12300?

Posted: **Sat Dec 29, 2007 9:00 pm**

for 12300 my output is:

32100 - 123 = 31977 = 9 * 3553

is it wrong?

32100 - 123 = 31977 = 9 * 3553

is it wrong?

Posted: **Sat Dec 29, 2007 9:06 pm**

yes, it is.S.M.ferdous wrote:for 12300 my output is:

32100 - 123 = 31977 = 9 * 3553

is it wrong?

all digits must occur in both a and b

Posted: **Sun Dec 30, 2007 1:39 am**

Hi, here is my code

I know it is a mess, but here are some tests I made:

I don't know what is wrong, I use short arrays so I don't have problems with large numbers.

THANKS

Sergio Ligregni, MEX

Code: Select all

```
/* Number Theory */
#include <stdio.h>
int rear (short [], short [], short);
int resta (short [], short [], short [], short);
int division (short [], short, short []);
int printe (short [], short [], short [], short []);
int main ()
{
/*FILE *in = freopen ("c.in", "r", stdin);/**/
short a[10];
short x=0;
short flag=1;
int v=0;
while (x++<10)
a[x-1] = 0;
short g=0;
while (flag)
{
/*
if (v!=0&&x!=0&&g!=10)
{
printf("\n");
}
v++;
*/
x=0;
while (x++<10)
a[x-1] = 0;
char c='\0';
x=0;
while ((c=getchar())!='\n'&&c!=EOF)
{
a[x] = c-'0';
x++;
}
if (c==EOF)
flag=0;
short g=0;
while (a[g]==0&&g<10)
g++;
if (x==0)
g==8;
if (x!=0&&g!=10)
{
/*
if (v!=0&&x!=0&&g!=9)
{
printf("\n");
}
v++;
*/
/*
printf("POR FAVOR!!!: %d %d\n", x, g);
*/
short f=9, temp[10], ind=0;
while (f>=0)
{
short r=g;
while (r<=9)
{
if (a[r]==f)
{
/*
printf("\n:%d g:%d x:%d r:%d\n", a[r], g, x, r);
*/
temp[ind] = f;
a[r] = -1;
ind++;
}
r++;
}
f--;
}
/* PRINTING *
short k=0;
while (k++<10)
printf(":%d", temp[k-1]);
printf("\n");
k=0;
while (k++<10)
printf(":%d", a[k-1]);
printf("\n");
/* EOP */
f=0;
while (f++<x-g)
a[f-1] = temp[f-1];
while (f++<10)
a[f-1] = 0;
short y=9, z=x-g-1;
while (y>=0)
{
if (z>=0)
{
a[y] = a[z];
z--;
}
else
a[y] = 0;
y--;
}
short b[10];
rear (a, b, x);
short residuo[10];
y=0;
while (y++<10)
residuo[y-1] = 0;
resta (a, b, residuo, x);
short cociente[10];
y=0;
while (y++<10)
cociente[y-1] = 0;
/*division (residuo, 9, cociente);*/
printe (a, b, residuo, cociente);
division (residuo, 9, cociente);
/*if (x!=0&&g!=9)
{
printf("\n");
}
v++;
*/
}
}
return 0;
}
int rear (short a[], short b[], short x)
{
short z=0, k=9, zeros=0, u=0;
while (a[k]==0)
{
zeros++;
k--;
}
b[z] = a[k];
k--;
z++;
while (u++<zeros)
{
b[z] = 0;
z++;
}
while (k>0)
{
b[z] = a[k];
k--;
z++;
}
/*
while (u++<zeros)
{
b[z] = 0;
z++;
}
*/
short y=9;
short j=0;
while (a[j]==0)
j++;
z=9-j;
while (y>=0)
{
if (z>=0)
{
b[y] = b[z];
z--;
}
else
b[y] = 0;
y--;
}
/* PRINTING *
k=0;
while (k++<10)
printf(":%d", a[k-1]);
printf("\n");
k=0;
while (k++<10)
printf(":%d", b[k-1]);
printf("\n");
/* EOP */
return 0;
}
int resta (short a[], short b[], short r[], short x)
{
short idiot=0, k=0;
while (k<10&&!idiot)
{
if (a[k]>b[k])
{
k = x+10;
}
else if (b[k]>a[k])
{
idiot = 1;
}
k++;
}
/* PRINTING *
k=0;
while (k++<10)
printf(":%d", a[k-1]);
printf("\n");
k=0;
while (k++<10)
printf(":%d", b[k-1]);
printf("\n");
/* EOP */
if (idiot)
{
k=0;
while (k<10)
{
short d=a[k];
a[k] = b[k];
b[k] = d;
k++;
}
}
/* PRINTING *
k=0;
while (k++<10)
printf(":%d", a[k-1]);
printf("\n");
k=0;
while (k++<10)
printf(":%d", b[k-1]);
printf("\n");
/* EOP */
k=0;
short p[10];
while (k++<10)
p[k-1] = b[k-1];
short pides=0;
short h=0;
h=9;
while (h>=0)
{
if (pides)
{
p[h]++;
pides = 0;
}
if (p[h]>a[h])
{
pides = 1;
r[h] = a[h]+10-p[h];
}
else
r[h] = a[h]-p[h];
h--;
}
return 0;
}
int division (short r[], short x, short c[])
{
unsigned long long num= 0;
short y=0;
while (y<10)
{
num = num*10 + r[y];
y++;
}
/*printf("%i\n", num/9L);*/
unsigned int div=0, res=0;
y=0;
short b=0;
while (div/9==0&&y<10)
{
div = div*10 + r[y];
y++;
}
if (y==10&&div==0)
printf("0");
if (div/9>0)
{
c[b] = div/9;
res = div%9;
b++;
}
if (y<10)
{
div = res*10+r[y];
y++;
while (div/9==0&&y<10)
{
div = div*10 + r[y];
y++;
c[b]=0;
b++;
}
if (div/9>0)
{
c[b] = div/9;
res = div%9;
b++;
}
}
if (y<10)
{
div = res*10+r[y];
y++;
while (div/9==0&&y<10)
{
div = div*10 + r[y];
c[b]=0;
b++;
y++;
}
if (div/9>0)
{
c[b] = div/9;
res = div%9;
b++;
}
}
if (y<10)
{
div = res*10+r[y];
y++;
while (div/9==0&&y<10)
{
div = div*10 + r[y];
c[b]=0;
b++;
y++;
}
if (div/9>0)
{
c[b] = div/9;
res = div%9;
b++;
}
}
if (y<10)
{
div = res*10+r[y];
y++;
while (div/9==0&&y<10)
{
div = div*10 + r[y];
c[b]=0;
b++;
y++;
}
if (div/9>0)
{
c[b] = div/9;
res = div%9;
b++;
}
}
if (y<10)
{
div = res*10+r[y];
y++;
while (div/9==0&&y<10)
{
c[b]=0;
b++;
div = div*10 + r[y];
y++;
}
if (div/9>0)
{
c[b] = div/9;
res = div%9;
b++;
}
}
if (y<10)
{
div = res*10+r[y];
y++;
while (div/9==0&&y<10)
{
div = div*10 + r[y];
c[b]=0;
b++;
y++;
}
if (div/9>0)
{
c[b] = div/9;
res = div%9;
b++;
}
}
if (y<10)
{
div = res*10+r[y];
y++;
while (div/9==0&&y<10)
{
div = div*10 + r[y];
c[b]=0;
b++;
y++;
}
if (div/9>0)
{
c[b] = div/9;
res = div%9;
b++;
}
}
if (y<10)
{
div = res*10+r[y];
y++;
while (div/9==0&&y<10)
{
div = div*10 + r[y];
c[b]=0;
b++;
y++;
}
if (div/9>0)
{
c[b] = div/9;
res = div%9;
b++;
}
}
if (y<10)
{
div = res*10+r[y];
y++;
while (div/9==0&&y<10)
{
div = div*10 + r[y];
c[b]=0;
b++;
y++;
}
if (div/9>0)
{
c[b] = div/9;
res = div%9;
b++;
}
}
if (y<10)
{
div = res*10+r[y];
y++;
while (div/9==0&&y<10)
{
div = div*10 + r[y];
c[b]=0;
b++;
y++;
}
if (div/9>0)
{
c[b] = div/9;
res = div%9;
b++;
}
}
short w=0;
while (w++<b)
printf("%d", c[w-1]);
/*printf("B: %d", b);*/
if (r[9]==0&&b!=0)
printf("0");
/**/
printf("\n");
/**/
return 0;
}
int printe (short a[], short b[], short r[], short c[])
{
short k=0;
while (a[k]==0)
k++;
while (k++<10)
printf("%d", a[k-1]);
printf(" - ");
k=0;
while (b[k]==0)
k++;
while (k++<10)
printf("%d", b[k-1]);
printf(" = ");
k=0;
while (r[k]==0&&k<9)
k++;
while (k++<10)
printf("%d", r[k-1]);
printf(" = 9 * ");
k=0;
while (c[k]==0)
k++;
while (k++<10)
printf("%d", c[k-1]);
k=0;
return 0;
}
```

Code: Select all

```
IN
12300
OUT
32100 - 10023 = 22077 = 9 * 2453
IN
00001
OUT
1 - 1 = 0 = 9 * 0
IN
10
OUT
10 - 10 = 0 = 9 * 0
IN
102
OUT
210 - 102 = 108 = 9 * 12
IN
1234567890
OUT
9876543210 - 1023456789 = 8853086428 = 9 * 983676269
```

THANKS

Sergio Ligregni, MEX

Posted: **Sun Dec 30, 2007 1:57 am**

You are using Big Integers, but actually long long int is enough.

With long long int, I think your code would be much shorter and easier to debug.

-----

Rio

With long long int, I think your code would be much shorter and easier to debug.

-----

Rio

Posted: **Sun Dec 30, 2007 3:12 am**

Thanks for your interest, but I'm a little scary about long long int, I used it in many problems and it didn't work,

so I started implementing short [], and I make the sustraction and division like elementary school,

I firstly wanna know I my tests are wrong, if there is a special number who could make wrong answers, and I could modify my code (I accept it is a terrible mess, it is in part because test printing).

Again, thanks!

Sergio Ligregni, MEX

so I started implementing short [], and I make the sustraction and division like elementary school,

I firstly wanna know I my tests are wrong, if there is a special number who could make wrong answers, and I could modify my code (I accept it is a terrible mess, it is in part because test printing).

Again, thanks!

Sergio Ligregni, MEX

Posted: **Sun Dec 30, 2007 3:16 am**

I just try your code with 1999999999 and it gives an obviously incorrect answer. I guess you are not initializing correctly?

Ah by the way, even if you feel uneasy with long long int, I think you may consider using double / long double.

Ah by the way, even if you feel uneasy with long long int, I think you may consider using double / long double.

Posted: **Sun Dec 30, 2007 4:27 am**

long long int doesn't work, I used double and got AC just now. Although n<=2000000000, how about 1999999999? the maximun of a-b is 9999999991 - 1999999999 = ... and the 9999999991 is greater than 2^64!rio wrote:You are using Big Integers, but actually long long int is enough.

With long long int, I think your code would be much shorter and easier to debug.

-----

Rio

Posted: **Sun Dec 30, 2007 4:41 am**

"...and the 9999999991 is greater than 2^64!" == FALSE

Posted: **Sun Dec 30, 2007 7:43 am**

How can I optimize my solution to run in <=1 sec? Here is my code:

Code: Select all

```
#define For(i,b) for(int i = 0; i < (int)b; ++i)
#define All(t) t.begin(),t.end()
string itos (int i) {stringstream s; s << i; return s.str();}
int stoi (string s) {istringstream in(s); int ret; in >> ret; return ret;}
string itos (int i) {stringstream s; s << i; return s.str();}
int stoi (string s) {istringstream in(s); int ret; in >> ret; return ret;}
int main ()
{
int n;
while(fin>>n)
{
string s=itos(n);
sort(All(s));
vector<string>v;
do
{
if(s[0]!='0')
v.push_back(s);
} while(next_permutation(All(s)));
int best=INT_MIN;
int a=0,b=0;
For(i,v.size())
{
Fori(j,i,v.size())
{
int aa=stoi(v[i]);
int bb=stoi(v[j]);
int x=abs(aa-bb);
if(x > best) {best=x;a=aa;b=bb;}
}
}
int at=a,bt=b;
a=max(at,bt);
b=min(at,bt);
int k=best/9;
printf("%d - %d = %d = 9 * %d\n",a,b,best,k);
}
return 0;
}
```

Posted: **Sun Dec 30, 2007 7:47 am**

sorry. you're right. But in C++, long long int seams to be 32-bits.sonyckson wrote:"...and the 9999999991 is greater than 2^64!" == FALSE

Posted: **Sun Dec 30, 2007 7:57 am**

Plz, try to find the exactly maximum of a-b! how about cin the n as string and sort it.armansuleimenov wrote:How can I optimize my solution to run in <=1 sec? Here is my code:

Code: Select all

`#define For(i,b) for(int i = 0; i < (int)b; ++i) #define All(t) t.begin(),t.end() string itos (int i) {stringstream s; s << i; return s.str();} int stoi (string s) {istringstream in(s); int ret; in >> ret; return ret;} string itos (int i) {stringstream s; s << i; return s.str();} int stoi (string s) {istringstream in(s); int ret; in >> ret; return ret;} int main () { int n; while(fin>>n) { string s=itos(n); sort(All(s)); vector<string>v; do { if(s[0]!='0') v.push_back(s); } while(next_permutation(All(s))); int best=INT_MIN; int a=0,b=0; For(i,v.size()) { Fori(j,i,v.size()) { int aa=stoi(v[i]); int bb=stoi(v[j]); int x=abs(aa-bb); if(x > best) {best=x;a=aa;b=bb;} } } int at=a,bt=b; a=max(at,bt); b=min(at,bt); int k=best/9; printf("%d - %d = %d = 9 * %d\n",a,b,best,k); } return 0; }`

Posted: **Sun Dec 30, 2007 8:43 am**

Hi!

I tried to make my code (Is the one all disordered above) with short [] (digit by digit) because I tried to use long long int before and I got WA, and with my rudimentary method (like child) I got AC (obviously it didn't help in the contest)

I wanna ask you something:

Type Bits

int 16

long 32

long long 64

So I tried (this problem specificly) with unsigned long, now I realize it is wrong because even the input is smaller, but when permuting, it can exceed the 4,xxx,xxx,xxx that unsigned long provides.

Someone suggested me to use**double**... but... can **double** offer me the precision I need???

Can**unsigned long long** (in theory up to 18,xxx,xxx,xxx,xxx,xxx,xxx) solve the problem?, does *unsigned* **long long** really exists? (I know is an stupid question, but here in my computer I tried to make calculations with 10-digit numbers with unsigned long long and it make wrong answers, but clearly, it could be my compiler, and UVa can accept that)

Thanks and sorry about my english!

Sergio Ligregni, MEX

I tried to make my code (Is the one all disordered above) with short [] (digit by digit) because I tried to use long long int before and I got WA, and with my rudimentary method (like child) I got AC (obviously it didn't help in the contest)

I wanna ask you something:

Type Bits

int 16

long 32

long long 64

So I tried (this problem specificly) with unsigned long, now I realize it is wrong because even the input is smaller, but when permuting, it can exceed the 4,xxx,xxx,xxx that unsigned long provides.

Someone suggested me to use

Can

Thanks and sorry about my english!

Sergio Ligregni, MEX

Posted: **Sun Dec 30, 2007 9:11 am**

that may be helpful.ligregni wrote:Hi!

I tried to make my code (Is the one all disordered above) with short [] (digit by digit) because I tried to use long long int before and I got WA, and with my rudimentary method (like child) I got AC (obviously it didn't help in the contest)

I wanna ask you something:

Type Bits

int 16

long 32

long long 64

So I tried (this problem specificly) with unsigned long, now I realize it is wrong because even the input is smaller, but when permuting, it can exceed the 4,xxx,xxx,xxx that unsigned long provides.

Someone suggested me to usedouble... but... candoubleoffer me the precision I need???

Canunsigned long long(in theory up to 18,xxx,xxx,xxx,xxx,xxx,xxx) solve the problem?, doesunsignedlong longreally exists? (I know is an stupid question, but here in my computer I tried to make calculations with 10-digit numbers with unsigned long long and it make wrong answers, but clearly, it could be my compiler, and UVa can accept that)

Thanks and sorry about my english!

Sergio Ligregni, MEX

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