HDU1719 Friend "Mathematical Law"

Topic Links:

http://acm.hdu.edu.cn/showproblem.php?pid=1719

Main topic:

Let's do a recursive definition of a friendly number:

(1) integers 1, 2 are friendly numbers

(2) If A and B are friendly numbers, then a*b + A + B is also a friendly number

(3) Only the numbers defined in (1) and (2) are friendly numbers.

Now give you a number n (0 <= N <= 30) to determine whether N is a friendly number

Ideas:

Very interesting a simple math problem. Set N is a friendly number, then n can be decomposed into:

n = a*b + A + b = (a+1) * (b+1)-1, i.e. n+1 = (a+1) * (b+1) is defined by recursion, where A and B must also be friendly numbers.

A = a1*b1 + a1 + B1 = (a1+1) * (b1+1)-1, i.e. A + 1 = (a1+1) * (b1+1), same as B + 1 = (a2+1) * (b2+1).

Substituting n + 1 = (a+1) * (b+1) is obtained: n + 1 = (a1+1) * (b1+1) * (a2+1) * (b2+1). And if only AI and bi are not

is 1 or 2, it can be broken down all the time. Finally get n + 1 = (1 + 1) ^x * (2 + 1) ^y = 2^x * 3^y.

This is good to do, first n plus 1, and then all 2 of the factor and 3 of the vegetarian factor all eliminated, if the elimination factor after the results are

1, the n+1 contains only 2 and 3 factors, then n is the friendly number. If the result of elimination factor after 1, that is, n also contains other factors,

Then n is not a friendly number.

AC Code:

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring>using namespace Std;int Main () {    int n;    Solve ();    while (~SCANF ("%d", &n))    {        if (n = = 0)        {            printf ("no!\n");            Continue;        }        n++;        while (n% 2 = = 0)            n/= 2;        while (n% 3 = = 0)            n/= 3;        if (n = = 1)            printf ("yes!\n");        else            printf ("no!\n");    }    return 0;}

HDU1719 Friend "Mathematical Law"