Ebony and Ivory_j - ebony and ivory-优快云博客

探讨一个关于游戏战斗中盾牌破坏的算法问题。主角需要通过使用两种不同枪支对盾牌造成特定数量的伤害来破坏它。文章提供了一个解决此问题的C语言实现方案。

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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.

Example

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.

其实这道题如果换成在数学里还是有点难（对我来说哈），但是用电脑就不一样了，因为你完全可以一个一个去试

题的大意就是问你，有没有这样一个整数，不，两个m与n，让m*x+n*y=sum，很简单吧。

好了，上代码~~~~~~~~~~~~~~

#include<stdio.h>
void f(int a,int b,int c)
{
    int i,k=0;
    for(i=0;i<=c/a;i++)
    {
        if((c-i*a)%b==0)
        {
            k=1;
            break;
        }
    }
    if(k==1)
    {
      printf("Yes\n");
    }
    else
    {
        printf("No\n");
    }
}
int main()
{
    int a,b,c,d;
    scanf("%d %d %d",&a,&b,&c);
    if(a>b)
    {
        d=a;
        a=b;
        b=d;
    }
    if(c<a)
    {
       printf("No\n");
    }
    else if(c<b&&c>=a)
    {
        if(c%a==0)
        {
            printf("Yes\n");
        }
        else
        {
            printf("No\n");
        }
    }
else if(c>=b)
    {
        f(a,b,c);
    }
    return 0;
}