CocktailWithSnake

NC222156. CocktailWithSnake

描述

The favorite game of Mr.Cocktail is dota2. Serpentine walking is a useful skill to dodge the enemy's attack. Serpentine walking refers to the path of movement that keeps swing direction like a snake. Suppose the map is a ${N \times M}$ two-dimensional plane. Mr.Cocktail at the point ${(1,1)}$ in initial.

He will first go to the right to the end, that is, point ${(N,1)}$ . Then turn and go one step upwards to point ${(N,2)}$ . Go to the left again to the end, that is, point ${(1,2)}$ ,. Then go one step up, and go to the right again to the end.

The picture above is the action line when ${N=3}$ and ${M=4}$ .

Now the map size is ${N \times M}$ . The number of steps ${k}$ of Cocktail's movement are known. Find the Manhattan distance from the position after ${k}$ steps to the starting point according to the rule.

Manhattan distance Definition: The distance between two points measured along axes at right angles. In a plane with ${p_1}$ at ${(x_1, y_1)}$ and $p_2$ at ${(x_2, y_2)}$ , it is $|x_1 - x_2| + |y_1 - y_2|$ .

输入描述

The first line contains a positive integer , which represents the number of groups to be tested. 

Next contains  lines, each row contains three integers

输出描述

For each test data, output a line of one coordinate to represent the answer. (A positive integer separated by two spaces represents the  and  coordinates respectively)

示例1

输入:

输出:

上次编辑到这里，代码来自缓存点击恢复默认模板

Python3 解法, 执行用时: 80ms, 内存消耗: 6876K, 提交时间: 2021-05-22 16:55:56

n = int(input())
nums = [list(map(int, input().split())) for _ in range(n)]
for i in range(n):
    a, b, c = nums[i][0], nums[i][1], nums[i][2]
    d = c // a
    if d & 1 == 1:
        print(d + a - 1 - (c % a))
    else:
        print(d + c % a)

C 解法, 执行用时: 13ms, 内存消耗: 456K, 提交时间: 2021-05-22 14:19:32

#include<stdio.h>
int main(void){
	long long k,x,y,t,n,m;
	scanf("%lld",&t);
	while(t--){
		scanf("%lld %lld %lld",&n,&m,&k);
		y=k/n;
		if(y%2==0)
			x=k%n;
		else
			x=n-1-k%n;
        printf("%lld\n",x+y);
	}
	return 0;
}

C++ 解法, 执行用时: 11ms, 内存消耗: 1144K, 提交时间: 2021-05-26 00:18:38

#include<stdio.h>
int main()
{
	long long k,x,y,t,n,m;
	scanf("%lld",&t);
	while(t--)
	{
		scanf("%lld %lld %lld",&n,&m,&k);
		y=k/n;
		if(y%2==0)
		x=k%n;
		else
		x=n-1-k%n;
		printf("%lld\n",x+y);
	}
	return 0;
}