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  • Manthan, Codefest 16 -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:判断c 能不能由A,B组成
    暴力枚举即可

    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <cstdlib>
    #include <string>
    #include <stack>
    #include <queue>
    #include <vector>
    #include <set>
    #include <list>
    #include <map>
    #include <iostream>
    #include <algorithm>
    
    using namespace std;
    
    typedef long long LL;
    
    const int INF = 0x3f3f3f3f;
    
    const double eps = 1e-6;
    
    const double PI = acos(-1.0);
    
    int a,b,c;
    
    int main()
    {
        cin>>a>>b>>c;
    
        bool flag=false;
        for(int i=0;a*i<=c;i++)
        {
            if((c-a*i)%b==0)
            {
                flag= true;
    
                break;
            }
        }
    
        if(flag)
        {
            cout<<"Yes"<<endl;
        }
        else
    
        {
            cout<<"No"<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/juechen/p/5255858.html
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