I just want to see if I say I will observe

x + y + z = 1

That is to say the X is less than 1 y is less than 1 Z is less than 1

It will be faster than

There are currently two more than 1 do not meet

Finally and most important point is that you'll find out it is equal to 1

And they are dozens of unexpectedly finally points equal to 1

You put them in a common denominator you'll find out that it is not hard to the molecule is equal to the denominator

x + y + z = 1

That is to say the X is less than 1 y is less than 1 Z is less than 1

It will be faster than

There are currently two more than 1 do not meet

Finally and most important point is that you'll find out it is equal to 1

And they are dozens of unexpectedly finally points equal to 1

You put them in a common denominator you'll find out that it is not hard to the molecule is equal to the denominator

Posted by Wendy at November 14, 2016 - 6:38 AM

I this is a stupid way

Look at the

a/(bc) +d/(ef) +g/(hi) =1

Among them:

bc = b*10+c

ef = e*10 + f

hi = h*10+i

Therefore:

a/(bc) +d/(ef) +g/(hi) =1

<==>

a*ef *hi +d *bc*hi + g* bc * ef = bc*ef*hi

The combination, then for each combination, check the a*ef *hi +d *bc*hi + g* BC * EF = = bc*ef*hi is established

Here are 2 bit integer calculation of *2 and *2, the result is 6 digits, not exceeding 32Bits.

Look at the

a/(bc) +d/(ef) +g/(hi) =1

Among them:

bc = b*10+c

ef = e*10 + f

hi = h*10+i

Therefore:

a/(bc) +d/(ef) +g/(hi) =1

<==>

a*ef *hi +d *bc*hi + g* bc * ef = bc*ef*hi

The combination, then for each combination, check the a*ef *hi +d *bc*hi + g* BC * EF = = bc*ef*hi is established

Here are 2 bit integer calculation of *2 and *2, the result is 6 digits, not exceeding 32Bits.

Posted by Jay at November 29, 2016 - 7:25 AM

The whole array

Posted by Hobart at December 11, 2016 - 7:43 AM

The 2 floor basic solution. . .

According to the 2 building ideas to write a small program, has unique solution (removal of exchange law)

According to the 2 building ideas to write a small program, has unique solution (removal of exchange law)

#include <stdio.h> #include <algorithm> #include <functional> using namespace std ; int main() { int a[] = {1,2,3,4,5,6,7,8,9}; do { int ef = a[4]*10+a[5]; int bc =a[1]*10+a[2]; int hi =a[7]*10+a[8]; if (a[0]*ef*hi+a[3]*bc*hi+a[6]*bc * ef ==bc*ef*hi ) { printf("%d/%d+%d/%d+%d/%d=1\n",a[0],bc,a[3],ef,a[6],hi); } }while (next_permutation(a,a+9)); getchar(); return 0; }

Posted by Katrina at December 17, 2016 - 8:16 AM

I will only use a stupid method,

Started by King at November 09, 2016 - 5:49 AM