题解归档 - cf2227A

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• 本地编号：cf2227A
• 本地来源：problems/cf2227A/idea.md
• 题目链接：https://codeforces.com/contest/2227/problem/A
• 原始标题：cf2227A - Koshary

思路

cf2227A - Koshary

pattern: parity reachability.

claim: long steps add 2 to one coordinate, and the single short step can fix the parity of at most one coordinate. Therefore (x, y) is reachable iff at most one of x, y is odd.

check: sample covers both odd (1 1, 5 9) and one/even parity cases.

代码

来源：problems/cf2227A/solution.cpp

/* Author: likely
 * Time: 2026-06-27
**/
#include<bits/stdc++.h>
#define ll long long
using namespace std;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin>>t;
    while(t--){
        int x,y;
        cin>>x>>y;
        cout<<(((x&1)+(y&1)<=1)?"YES":"NO")<<"\n";
    }
    return 0;
}