Codeforces 633A Ebony and Ivory 【水题】

A. Ebony and Ivory

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.

For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units
of damage. In order to break the shield Dante has to deal exactly c units
of damage. Find out if this is possible.

Input

The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) —
the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.

Output

Print "Yes" (without quotes) if Dante can deal exactly c damage
to the shield and "No" (without quotes) otherwise.

Examples

input

4 6 15

output

No

input

3 2 7

output

Yes

input

6 11 6

output

Yes

Note

In the second sample, Dante can fire 1 bullet from Ebony and 2 from
Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet
from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.

题意：给你三个数a、b、c，问你存不存在一种方案使得a*x + b*y = c。x、y为非负数。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;
int main()
{
int a, b, c;
cin >> a >> b >> c; bool flag = false;
for(int i = 0; i <= c / a; i++)
{
if((c - i*a) % b == 0)
{
flag = true;
break;
}
}
printf(flag ? "Yes\n" : "No\n");
return 0;
}