zoukankan      html  css  js  c++  java
  • Codeforces Round #624 (Div. 3) A. Add Odd or Subtract Even(水题)

    You are given two positive integers aa and bb .

    In one move, you can change aa in the following way:

    • Choose any positive odd integer xx (x>0x>0 ) and replace aa with a+xa+x ;
    • choose any positive even integer yy (y>0y>0 ) and replace aa with aya−y .

    You can perform as many such operations as you want. You can choose the same numbers xx and yy in different moves.

    Your task is to find the minimum number of moves required to obtain bb from aa . It is guaranteed that you can always obtain bb from aa .

    You have to answer tt independent test cases.

    Input

    The first line of the input contains one integer tt (1t1041≤t≤104 ) — the number of test cases.

    Then tt test cases follow. Each test case is given as two space-separated integers aa and bb (1a,b1091≤a,b≤109 ).

    Output

    For each test case, print the answer — the minimum number of moves required to obtain bb from aa if you can perform any number of moves described in the problem statement. It is guaranteed that you can always obtain bb from aa .

    Example
    Input
     
    5
    2 3
    10 10
    2 4
    7 4
    9 3
    
    Output
     
    1
    0
    2
    2
    1
    判断几次就行了。
    #include <bits/stdc++.h>
    using namespace std;
    int main()
    {
        int t;
        cin>>t;
        while(t--)
        {
            int a,b;
            cin>>a>>b;
            if(b>a)
            {
                if((b-a)%2!=0)
                {
                    cout<<1<<endl;
                    continue;
                }
                else
                {
                    cout<<2<<endl;
                    continue;
                }
            } 
            else if(b==a)
            {
                cout<<0<<endl;
                continue;
            }
            else
            {
                if((a-b)%2!=0)
                {
                    cout<<2<<endl;//a先加上1变成偶数 
                    continue;
                }
                else cout<<1<<endl;
                continue;
            }
        }
        return 0;
    }
  • 相关阅读:
    其他
    聚类算法:ISODATA算法
    大神博客
    Fiddldr 教程之:HTTP协议详解(转)
    设计模式之装饰模式的复习
    NOIP 2011 聪明的质监员
    CSP-S2020/NOIP2020模板总结(Updating)
    CSP-S2020/NOIP2020复习指南
    洛谷 U137412 高斯的小宇宙
    NOIP2020模板测试题大全
  • 原文地址:https://www.cnblogs.com/lipoicyclic/p/12372181.html
Copyright © 2011-2022 走看看