zoukankan      html  css  js  c++  java
  • Codeforces Round #597 (Div. 2) A. Good ol' Numbers Coloring

    链接:

    https://codeforces.com/contest/1245/problem/A

    题意:

    Consider the set of all nonnegative integers: 0,1,2,…. Given two integers a and b (1≤a,b≤104). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.

    Each number is painted white or black. We paint a number i according to the following rules:

    if i=0, it is colored white;
    if i≥a and i−a is colored white, i is also colored white;
    if i≥b and i−b is colored white, i is also colored white;
    if i is still not colored white, it is colored black.
    In this way, each nonnegative integer gets one of two colors.

    For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...

    Note that:

    It is possible that there are infinitely many nonnegative integers colored black. For example, if a=10 and b=10, then only 0,10,20,30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
    It is also possible that there are only finitely many nonnegative integers colored black. For example, when a=1 and b=10, then there is no nonnegative integer colored black at all.
    Your task is to determine whether or not the number of nonnegative integers colored black is infinite.

    If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).

    思路:

    猜的GCD, 考虑ax+by = gcd(a, b),所以要整除。

    代码:

    #include<bits/stdc++.h>
    using namespace std;
     
    int main()
    {
        ios::sync_with_stdio(false);
        int t;
        cin >> t;
        while(t--)
        {
            int a, b;
            cin >> a >> b;
            if (__gcd(a, b) != 1)
                puts("Infinite");
            else
                puts("Finite");
        }
     
        return 0;
    }
    
  • 相关阅读:
    having——至少被订购过两回的订单
    产品——仓库表查询
    SQL 聚集函数使用
    select count(*)和select count(1)的区别 (转)
    SpringAOP 通知(advice)
    Spring AOP 实现原理与 CGLIB 应用
    cglib 动态代理基础篇
    模仿Struts2的Interceptor拦截器实现
    利用JDK动态代理机制实现简单拦截器
    java多线程总结二:后台线程(守护线程)
  • 原文地址:https://www.cnblogs.com/YDDDD/p/11797722.html
Copyright © 2011-2022 走看看