马的走法C语言算法

[C语言数组排序]#include<stdio.h> void main() { int a[10] = { 10,2,3,4,5,6,9,8,7,1 }; int i,j,t; for(j=0;j<10;j++) for(i=0;i<10-1-j;i++) if(a[i]>a[i+1]) /* 由小到大，由大到小时改...+阅读

#include

* （x1,y1）向右走到（y1,y2）可行的走法总数

int getWays(int x1, int y1, int x2, int y2)

{

if(y1 > y2) //在目标右边

return 0;

else if(y1 == y2 & x1 != x2) //在目标一条垂直线上，但不是同一点

return 0;

else if(y1 == y2 & x1 == x2) //跟目标是同一点

return 1;

else{ //分治，四种走法的总和

int result =0;

if(x1 > 1) //可以上1

result += getWays(x1 - 1, y1 + 2, x2, y2);

if(x1 > 2) //可以上2

result += getWays(x1 - 2, y1 + 1, x2, y2);

if(x1 < 5) //可以下1

result += getWays(x1 + 1, y1 + 2, x2, y2);

if(x1 < 4) //可以下2

result += getWays(x1 + 2, y1 + 1, x2, y2);

return result;

}

void main(void)

{

int x,y； //目标坐标

printf（＂请输入目标坐标：\n＂）;

scanf(＂%d%d＂, &x, &y);

printf（＂总的不同走法：%d\n＂,getWays(1, 1, x, y));

}

不明白，再交流...

棋盘覆盖算法

import java.util.*; public class TestChessBoard { public static void main(String[] args) { int tr=0,tc=0,dr=1,dc=2,size=8; ChessBoard.chessBoard(tr,tc,dr,dc,size); ChessBoard.display(); } } class ChessBoard { public static int tile = 0; public static int[][] board= new int[10][10]; public static void chessBoard (int tr,int tc,int dr,int dc,int size) { if(size == 1) return; int t = tile++ , s = size/2; if(dr=tc+s){ chessBoard(tr,tc+s,dr,dc,s); }else { board[tr+s-1][tc+s] = t; chessBoard(tr,tc+s,tr+s-1,tc+s,s); } if(dr>=tr+s & dc=tr+s & dc>=tc+s) { chessBoard(tr+s,tc+s,dr,dc,s); }else { board[tr+s][tc+s] = t; chessBoard(tr+s,tc+s,tr+s,tc+s,s); } } public static void display() { for(int i=0;i

棋盘覆盖问题算法 pree

var a,b,c:array[1..100000] of longint; n,i,j,y,t:longint;begin readln(n); a[1]:=1; b[1]:=2; if n=1 then begin writeln

(1); halt; end; if n=2 then begin writeln

(2); halt; end; t:=1; for i:=3 to n do begin y:=0; for j:=1 to t do begin c[j]:=b[j]+a[j]+y; y:=c[j] div 10; c[j]:=c[j] mod 10; end; if y0 then begin inc(t); c[t]:=y; end; a:=b; b:=c; fillchar(c,sizeof(c),0); end; for i:=t downto 1 do write(b[i]); writeln;end....