OJ Board

hi osan still now i got wrong ans.i change my source code what u tell me.if i give input 999999 & 999999 give output 1.but if i give input 555555&555555 || 666666&666666 || 777777&777777 || 888888&888888 give correct ans.

hi everybody at last i solve this problem.i only change char t[30000].

[c]

Firstly sum up all the digit of the number then sum up the factorizing
number.if sum of each digit is equal to the sum of factorizing number
then the number is called smith number.

sample input:
-------------

20
4937774
4997775
5445454
4648787
4897987
5789798
9554564
8545644
7564545
8798798 ...

[c]

#include<iostream.h>
#include<string.h>
#include<math.h>
#include<stdio.h>
void main()
{
unsigned long int n,check=0,l,m,che=0,b,ch=0,t,j;
unsigned long int array[3000]={0};
while(cin>>m)
{
for(l=0;l<m;l++)
{
cin>>array[l];
}
b=abs(array[1])-abs(array[0]);


for(j=1;j<m-1;j++)
{
if(abs ...

[c]
this is big input problem.i use following rule.about all of the input
output is ok.but some input output is not match in actual input output.help me.



any number divisible by two if last digit is divisible by two.
any number divisible by 3 if sum of digits is divisible by 3.
any number ...

[c]
#include<stdio.h>
#include<string.h>
#include<ctype.h>
#define max 30000

void main()
{
char x1[max],x4[max];
char *t,*s,c;
int i=0,p=0,q=0,l=0,b=0,check=0,che=0,y;
while(gets(x1))
{
l=strlen(x1);
t=x1;
s=x1;
while(*t)
{
memset(x4,'\0',max);
s=s+q;
q=0;
b=0;
che=0;
check=0;
while(*t!=' '&&*t ...

[c]
My program gives following output for the below input.but i dont understand
why wrong ans.plz check my solution.I think a little problem is arise.
help me.

input::
--------
123 456
555 555
123 594
0 1
1 0
999999999 1
1 999999999
111111111 888888888
111111112 888888888
900000000 100000000
1 1
1 ...

[c]
PE help me plz.here is my source code.

/* @JUDGE_ID: xxxxxx 160 C++ */

#include<stdio.h>
#include<string.h>
int main()
{
long int i,j,n,k,p=0;
static int x[500];
memset(x,0,500);
scanf("%ld",&n);
while(n>0)
{ for(i=2;i<=n;i++)
{ j=i;
for(k=2;k<=i;k++)
{ while(j%k==0)
{ ++x[k];
j/=k ...

PE help me plz.here is my source code.

/* @JUDGE_ID: xxxxxx 160 C++ */

#include<stdio.h>
#include<string.h>
int main()
{
long int i,j,n,k,p=0;
static int x[500];
memset(x,0,500);
scanf("%ld",&n);
while(n>0)
{ for(i=2;i<=n;i++)
{ j=i;
for(k=2;k<=i;k++)
{ while(j%k==0)
{ ++x[k];
j/=k ...

[c]
I use this formula.but why TLE.plz help me.my source code is in above.
i increase array size to 1000000.

a1=(a0+a2)/2-c1
2a1=a0+a2-2c1
4a1=2a0+2a2-4c1 ... (1)

a2=(a1+a3)/2-c2
2a2=a1+a3-2c2 ... (2)

Combine (1) & (2)

4a1=2a0+(a1+a3-2c2)-4c1
3a1=2a0+a3-2(c2+2c1)

... and so on until n-th term ...

seems very simple.but wa.i m frustrated.plz rescue me.

[c]

#include<stdio.h>
#include<math.h>
#include<string.h>
#define max 10000
void main()
{
char t[max];
long int i,j,l;
scanf("%ld",&j);
for(l=0;l<=j;l++)
{
gets(t);
i=strlen(t);
if(t[0]=='1'&&t[1]=='9'&&t[2]=='0')
printf("?\n");
if(t[i-1]=='5 ...

here is my code.wa.plz help me.
[c]

#include<stdio.h>
#include<math.h>

#define SIZE 5
#define SIZE1 7500



void main()
{
long int store[SIZE]={1,5,10,25,50};
long int count[SIZE1],inp;
long int i,j,temp;

while(scanf("%ld",&inp)==1)
{

if(inp==0)
{
printf("There is only 1 way to produce %ld cents ...

why wa plz help me.here is my code.
[c]
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<ctype.h>

binary(int sum)
{
int t[40];
int i=0,c;
memset(t,0,30);
while(sum>0)
{
t[i++]=fmod(sum,2)+48;
sum=sum/2;
}
c=13-i-1;
while(c>=0)
{
printf("0");
c--;
}
for(i=i-1;i>=0;i--)
printf("%c",t ...

TLE plz give me correct solution.


[c]

#include<stdio.h>
#include<math.h>


void main()
{
int i,n,j,k;
double a[10000]={0.0};
double sum=0.0;
while(scanf("%d",&i)!=EOF)
{
for(j=0;j<n;j++)
{
scanf("%d",&n);
for(k=0;k<n+2;k++)
scanf("%lf",&a[k]);
for(k=0;k<n;k++)
sum=(sum+(a[k+2]*(n-k)));
sum=((a[0 ...

wa.think its very simple.every input output is ok.plz help me.

[cpp]

/* @JUDGE_ID: xxxxxx 160 C++ */


#include<stdio.h>
int main()
{
long int i,j,n,k,p=0;
static int x[100];
scanf("%ld",&n);
while(n>0)
{ for(i=2;i<=n;i++)
{ j=i;
for(k=2;k<=i;k++)
{ while(j%k==0)
{ ++x[k];
j/=k ...