976 B. Lara Croft and the New Game

最新推荐文章于 2020-12-17 08:23:13 发布

weixin_30677617

最新推荐文章于 2020-12-17 08:23:13 发布

阅读量210

点赞数

CC 4.0 BY-SA版权

文章标签：开发工具

原文链接：http://www.cnblogs.com/shinianhuanniyijuhaojiubujian/p/8977195.html

本文介绍了一款Lara Croft系列游戏中关于探索地下城的路径算法。通过输入地下城的尺寸和Lara移动次数，可以计算出她的当前位置。文章提供了一个C++实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

You might have heard about the next game in Lara Croft series coming out this year. You also might have watched its trailer. Though you definitely missed the main idea about its plot, so let me lift the veil of secrecy.

Lara is going to explore yet another dangerous dungeon. Game designers decided to use good old 2D environment. The dungeon can be represented as a rectangle matrix of n rows and m columns. Cell (x, y) is the cell in the x-th row in the y-th column. Lara can move between the neighbouring by side cells in all four directions.

Moreover, she has even chosen the path for herself to avoid all the traps. She enters the dungeon in cell (1, 1), that is top left corner of the matrix. Then she goes down all the way to cell (n, 1) — the bottom left corner. Then she starts moving in the snake fashion — all the way to the right, one cell up, then to the left to the cell in 2-nd column, one cell up. She moves until she runs out of non-visited cells. n and m given are such that she always end up in cell (1, 2).

Lara has already moved to a neighbouring cell k times. Can you determine her current position?

Input

The only line contains three integers n, m and k (2 ≤ n, m ≤ 10⁹, n is always even, 0 ≤ k < n·m). Note that k doesn't fit into 32-bit integer type!

Output

Print the cell (the row and the column where the cell is situated) where Lara ends up after she moves k times.

Examples

Input

Copy

4 3 0

Output

Copy

1 1

Input

Copy

4 3 11

Output

Copy

1 2

Input

Copy

4 3 7

Output

Copy

3 2

Note

Here is her path on matrix 4 by 3:

$\text{[math]}$

找规律

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 0x3f3f3f3f
#define mem(a) ((a,0,sizeof(a)))
typedef long long ll;
ll n,m,k;
int main()
{
    scanf("%lld%lld%lld",&n,&m,&k);
    if(k<=(n-1)) printf("%lld 1\n",1+k);
    else
    {
        k-=n-1;
        ll ans=k/(m-1);
        ll pos=k%(m-1);
        if(pos==0) printf("%lld %lld\n",n-ans+1,ans&1?m:2);
        else printf("%lld %lld\n",n-ans,ans&1?m+1-pos:1+pos);
    }
    return 0;
}