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вернуться в форумi've got AC by this program(inside) but it take 3 sec .Can anybody give me some hint to solve it in 0.02 sec #include <stdio.h>
int main()
{
long n,limit,i;
scanf("%ld",&n);
limit=n/2;
for(i=3;i<=limit;i++)
if(n%i==0) { n=i; break; }
printf("%ld",n-1);
return 0;
} By Maths, you only need to check to the square root of limit(-) > #include <stdio.h>
> int main()
> {
> long n,limit,i;
> scanf("%ld",&n);
> limit=n/2;
> for(i=3;i<=limit;i++)
> if(n%i==0) { n=i; break; }
> printf("%ld",n-1);
> return 0;
> } but if n=2*(prime number) eg. n=26 and another case > > #include <stdio.h>
> > int main()
> > {
> > long n,limit,i;
> > scanf("%ld",&n);
> > limit=n/2;
> > for(i=3;i<=limit;i++)
> > if(n%i==0) { n=i; break; }
> > printf("%ld",n-1);
> > return 0;
> > } So, two special cases arise :) (-) > > > #include <stdio.h>
> > > int main()
> > > {
> > > long n,limit,i;
> > > scanf("%ld",&n);
> > > limit=n/2;
> > > for(i=3;i<=limit;i++)
> > > if(n%i==0) { n=i; break; }
> > > printf("%ld",n-1);
> > > return 0;
> > > } Thank you very much for ur help :-D > > > > #include <stdio.h>
> > > > int main()
> > > > {
> > > > long n,limit,i;
> > > > scanf("%ld",&n);
> > > > limit=n/2;
> > > > for(i=3;i<=limit;i++)
> > > > if(n%i==0) { n=i; break; }
> > > > printf("%ld",n-1);
> > > > return 0;
> > > > } |
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