Diagonal Walking v.2(思维)
Mikhail walks on a Cartesian plane. He starts at the point (0,0)(0,0), and in one move he can go to any of eight adjacent points. For example, if Mikhail is currently at the point (0,0)(0,0), he can go to any of the following points in one move:
(1,0)(1,0);
(1,1)(1,1);
(0,1)(0,1);
(−1,1)(−1,1);
(−1,0)(−1,0);
(−1,−1)(−1,−1);
(0,−1)(0,−1);
(1,−1)(1,−1).
If Mikhail goes from the point (x1,y1)(x1,y1) to the point (x2,y2)(x2,y2) in one move, and x1≠x2x1≠x2 and y1≠y2y1≠y2, then such a move is called a diagonal move.
Mikhail has qq queries. For the ii-th query Mikhail’s target is to go to the point (ni,mi)(ni,mi) from the point (0,0)(0,0) in exactly kiki moves. Among all possible movements he want to choose one with the maximum number of diagonal moves. Your task is to find the maximum number of diagonal moves or find that it is impossible to go from the point (0,0)(0,0) to the point (ni,mi)(ni,mi) in kiki moves.
Note that Mikhail can visit any point any number of times (even the destination point!).
Input
The first line of the input contains one integer qq (1≤q≤1041≤q≤104) — the number of queries.
Then qq lines follow. The ii-th of these qq lines contains three integers nini, mimi and kiki (1≤ni,mi,ki≤10181≤ni,mi,ki≤1018) — xx-coordinate of the destination point of the query, yy-coordinate of the destination point of the query and the number of moves in the query, correspondingly.
Output
Print qq integers. The ii-th integer should be equal to -1 if Mikhail cannot go from the point (0,0)(0,0) to the point (ni,mi)(ni,mi) in exactly kiki moves described above. Otherwise the ii-th integer should be equal to the the maximum number of diagonal moves among all possible movements.
Example
Input
3
2 2 3
4 3 7
10 1 9
Output
1
6
-1
Note
One of the possible answers to the first test case: (0,0)→(1,0)→(1,1)→(2,2)(0,0)→(1,0)→(1,1)→(2,2).
One of the possible answers to the second test case: (0,0)→(0,1)→(1,2)→(0,3)→(1,4)→(2,3)→(3,2)→(4,3)(0,0)→(0,1)→(1,2)→(0,3)→(1,4)→(2,3)→(3,2)→(4,3).
In the third test case Mikhail cannot reach the point (10,1)(10,1) in 9 moves.
被这个题卡了好久,最后还是看题解想出来的.画图的话各种情况是都要考虑到的.不过说到底还是考虑各种奇偶性.还有就是k的定义.我开始以为k是最多要走的.但不是必须要走的.后来发现就是必须要走的.!!!这是个坑,否则就随便走就行了…
代码如下:
#include
#include
#include
#include
#include
#define ll long long
using namespace std;
ll n,m,k;
int main()
{
int t;
cin>>t;
while(t--)
{
scanf("%I64d%I64d%I64d",&n,&m,&k);
if(n>m) swap(n,m);//比较矩形的长
if(k 努力加油a啊,(o)/~