codeforces cf edu round#50 B. Diagonal Walking v.2

Posted 2021-01-31 lishengkangshidatiancai

思路：当m > k时输出-1（设m是较大的数），当m-n是奇数时有一步不能走对角线所以k--，当走对角线可以直接到达终点，如果剩余的步数是奇数则有两步不能走对角线所以k - 2。（画图观察规律）

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
	int q;
	scanf("%d",&q);
	while(q--)
	{
		ll n,m,k;
		scanf("%lld%lld%lld",&n,&m,&k);
		if(n>m)
		  swap(n,m);
		if(k<m)
		{
			printf("-1
");
			continue;
		}
		ll res;
		if( (m-n)%2==0 )
		{
			if(k%2==m%2)
			{
				printf("%lld
",k);
			}
			else
			{
				printf("%lld
",k-2);
			}
		}
		else if( (m-n)%2==1 ) 
		{
			if(k==m)
			{
				printf("%lld
",k-1);
				continue;
			}
			else if(k>m)
			{
			    ll sheng=k-n;
			    ll cha=m-n;
			    if(sheng%2==cha%2)
			    {
			    	if(cha%2==0)
			    	{
			    		res=k;
					}
					else if(cha%2==1)
					{
						res=k-1;
					}
				}
				else
				{
					if(cha%2==1&&sheng%2==0)
					{
						res=k-1;
					}
					else if(cha%2==0&&sheng%2==1)
					{
						res=k-2;
					}
				}
				printf("%lld
",res);
				continue;
			}
		}
	}
}