codeforces 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a  b  c三个数，问是否存在x  y使得a*x+b*y=c

数据不大直接暴力求解

#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD double
#define MAX 100100
#define mod 10007
#define dian 1.000000011
using namespace std;
int main()
{
    int a,b,c,t,n,m,i,j;
    while(scanf("%d%d%d",&a,&b,&c)!=EOF)
    {
    	int flag=0;
    	for(i=0;i<=c/a;i++)
    	{
    	    if((c-i*a)%b==0)
    	    {
    	    	flag=1;
    	    	break;
    	    }
    	}
    	if(flag) printf("Yes\n");
    	else printf("No\n");
    }
	return 0;
}