Ural 1430. Crime and Punishment

1430. Crime and Punishment

Time limit: 0.5 second
Memory limit: 64 MB

Background

Petty bureaucrat Victor Thiefton was disposed towards stealing from his childhood. But one thing is to legally privatize national factories, diamond fields and oil derricks at the cost of billions dollars. And another thing is to filch some money from a poor regional budget. Our legislation is very strict. Therefore Victor felt that justice is on the alert just after he extracted his hand from the national pocket. What should he do to escape inevitable punishment?

Mr. Thiefton has once heard that in accordance with the criminal legislation standards he would be condemned to long imprisonment for a theft whereas in case of a peculation he could escape with a suspended sentence only. So if the most part of stolen money is peculated, the duration of imprisonment will be reduced.

Problem

The same evening Mr. Thiefton burst into "MegaApril" superstore and rushed for overflowing storefronts carrying a purse with N stolen dollars. It appeared that unlimited number of high-quality goods and goods at moderate price were on sale in the superstore. High-quality goods cost A dollars per piece, and goods at moderate price cost B dollars per piece. Victor should spend as much stolen money as possible to reduce the duration of imprisonment to a minimum.

Input

The only line contains the integer numbers A, B and N (1 ≤ A, B, N ≤ 2∙109).

Output

You should output the number of high-quality goods and the number of goods at moderate price, which should be bought to guarantee the minimal duration of imprisonment for Victor. The numbers should be separated by single space. If the problem has several solutions, you should output any of them.

Sample

inputoutput

8 5 22

2 1

Problem Author: Nikita Rybak, Ilya Grebnov, Dmitry Kovalioff
Problem Source: Timus Top Coders: First Challenge

用n元买价值分别为a，b的东西，要使剩余的钱最少。

暴力就能过

#include<iostream>using namespace std;#define MIN 0x7ffffffflong long  a,b,n,t,x,d,i;bool flag;int main(){    cin>>a>>b>>n;    x=0;    if (a < b)    {        swap(a,b);        flag=true;    }    t=min(n/a,b);    d=MIN;    for (i=0; i<=t; i++)        if ((n-a*i)%b < d)        {            d=(n-a*i)%b;            x=i;        }    if (!flag)        cout << x << ' ' << (n-a*x)/b << endl;    else        cout << (n-a*x)/b << ' ' << x << endl;    return 0;}