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  • 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 abc (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
    ");
        	else printf("No
    ");
        }
    	return 0;
    } 
    

      

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  • 原文地址:https://www.cnblogs.com/tonghao/p/5230849.html
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