UVa 11371 number theory for newbies (water ver.)

11371-number Theory for Newbies

Time limit:1.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem &problem=2366

Given any positive integer, if we permute it digits, the difference between the number we get and the Given number would a Lways is divisible by 9. For example, if the given number is 123, we'll rearrange the digits to get 321. The difference = 321-123 = 198, which is a multiple of 9 (198 = 9x22).

We can prove this fact fairly easily, but since we are do not have a maths contest, we instead try to illustrate this fact With the help of a computer program.

Input and Output

Each line of input gives a positive integer n (≤2000000000). You are to find two Integersa andb formed by rearranging the digits ofn, such thata-b is MAXIMUM.A andb-should not have LE Ading Zeros. You are should then show Thata-b are a multiple of 9, by expressing it as 9XK, and Wherek is an integer. The sample output for the correct output format.

Sample Input

123
2468

Sample Output

321-123 = 198 = 9 *
8642-2468 = 6174 = 9 * 686

Long long, grave-Egg!

Complete code:

/*0.015s*/
  
#include <bits/stdc++.h>
using namespace std;
  
Char a[15], b[15];
  
int main ()
{
    int len, I;
    Long long diff;
    while (gets (a))
    {
        len = strlen (a);
        Sort (A, a + Len, greater<char> ());
        memcpy (b, A, sizeof (a));///This is b for
        (i = 0; i < len; ++i) A[i] = b[len-1-i];
        for (i = 0;!) ( A[i] & 15); ++i)
            ;
        Swap (a[0], a[i]);
        diff = Atoll (b)-Atoll (a);
        printf ("%s-%s =%LLD = 9 *%lld\n", B, A, diff, diff/9);
    }
    return 0;
}

Author: csdn Blog Synapse7

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