POJ 1728 A flea on a chessboard

最新推荐文章于 2019-04-17 23:23:03 发布

原创最新推荐文章于 2019-04-17 23:23:03 发布 · 1k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#input #numbers #output #each #up

数论专栏收录该内容

15 篇文章

订阅专栏

本文探讨了一个位于无限延伸的黑白相间棋盘上的跳蚤问题。通过输入跳蚤的初始位置及跳跃方向，程序判断跳蚤跳到白色方格所需的最少跳跃次数，或判断其是否永远无法跳出黑色方格。

A flea on a chessboard

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 1509		Accepted: 491

Description

An infinite chessboard is obtained by extending a finite chessboard to the right and up infinitely. Each square of the chessboard is either black or white with the side of S milimiters, 0 < S <= 1000. The leftmost bottom square of the chessboard is black. A flea is positioned on the chessboard at the point (x, y) (given in milimeters) and makes jumps by jumping dx milimeters to the right and dy milimiters up, 0 < dx, dy, that is, a flea at position (x, y) after one jump lands at position (x+dx, y+dy).
Given the starting position of the flea on the board your task is to find out after how many jumps the flea will reach a white square. If the flea lands on a boundary between two squares then it does not count as landing on the white square. Note that it is possible that the flea never reaches a white square.

Input

Each test case consists of one line of input containing five non-negative numbers separated by white space and giving integers S, x, y, dx, and dy. An input line containing five zeroes follows the last test case.

Output

For test case print one line of output in the format shown in the sample.

Sample Input

10 2 3 3 2
100 49 73 214 38
25 0 0 5 25
407 1270 1323 1 1
18 72 6 18 6
407 1270 1170 100 114
0 0 0 0 0

Sample Output

After 3 jumps the flea lands at (11, 9).
After 1 jumps the flea lands at (263, 111).
The flea cannot escape from black squares.
After 306 jumps the flea lands at (1576, 1629).
The flea cannot escape from black squares.
After 0 jumps the flea lands at (1270, 1170).

Source

Waterloo local 2004.01.31

/* 首先很容易判断出当前坐标(x, y)落在白格子上的条件是q(x / S) + q(y / S) 为奇数, q表示下取整那么可以从起点不断地循环跳,并用上述条件判断是否跳到了白格子, 如果不在白格子，则对当前当前坐标模S，当(x % S, y % S)已经被访问过时说明无解这里有一个要注意的地方，这个地方让我A了很多次，那就是: 模运算时如果x % 2 == 0(或者y % 2 == 0)时即落在边界上，需要判断这个点是在格子最左侧边界还是在右侧边界，判断的方法是判断当前这个格子的颜色，如果是黑格子则为起始处，如果是白格子则是在最右侧 */ #include <iostream> #define MAX_N 1005 using namespace std; bool v[MAX_N + 1][MAX_N + 1]; int main() { __int64 S, x, y, dx, dy; while(scanf("%I64d%I64d%I64d%I64d%I64d", &S, &x, &y, &dx, &dy) && (S + x + y + dx + dy) != 0) { memset(v, 0, sizeof(v)); __int64 curX = x, curY = y, steps = 0; bool can = true; while(true) { __int64 r1 = curX % S, r2 = curY % S; __int64 q1 = curX / S, q2 = curY / S; if(r1 != 0 && r2 != 0 && (q1 + q2) % 2 == 1) break; //判断在哪个边界 if(r1 == 0 && (q1 + q2) % 2 == 1) r1 = S; if(r2 == 0 && (q1 + q2) % 2 == 1) r2 = S; if(v[r1][r2]) { can = false; break; } steps++; v[r1][r2] = true; curX += dx; curY += dy; } if(!can) printf("The flea cannot escape from black squares./n"); else printf("After %I64d jumps the flea lands at (%I64d, %I64d)./n", steps, curX, curY); } return 0; }