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pAsCAl算24点

const as:array[1..24] of integer=(1234,1243,1324,1342,1423,1432,2134,2143,2314,2341,2413,2431,3124,3142,3214,3241,3412,3421,4123,4132,4213,4231,4312,4321);{穷举所有可能的排列} var a,b,c,d,i,j,k,l,t,t1,t2,o:integer; {t存放每次的...

Type nt=record val:real; c:integer; s:string[100]; end; an=array[1..4] of nt; Var b:an; a:array[1..4] of real; i:integer; found,error:boolean; Function ist(n:integer):string; var s:string; begin s:=''; repeat s:=chr(ord('0')+n ...

type arr=array[1..4] of integer; var d:arr; f:array[1..3,1..4] of integer; i,j,m,n:integer; procedure print; var i,j:integer; begin for i:=1 to 3 do begin for j:=1 to 3 do if j2 then write(f[i,j]) //除2以外其余都是数字 else cas...

可能有点小超时(pp为个数,ppp为所求数) Type nt=record val:real; c:integer; s:string[100]; end; an=array[1..100] of nt; Var b:an; a:array[1..100] of real; pp,ppp,i:integer; found,error:boolean; Function ist(n:integer):string; va...

程序如下: var a:array[1..2000,0..4]of integer; b:array[1..2000]of integer; c:array[1..2000,1..2]of integer; d:array[1..2000]of char; i,j,x,y,m,n,k,l,t:integer; procedure print(j:integer); var f:integer; begin if j=0 then exit;...

回溯

下面是N年前用JS写的,那时刚学编辑,方法很笨,连命名都不规范,后面写了一个很简洁的,但代码找不见了,和pascal语法不一样,你可以参考一下编辑思路,然后改写一下 function fun1(e1,e2,e3,e4) { var i,j,k,a,b,c,d,t1,t2,t3; var opx=["+","...

program E1_1; {knight} const dx:array[1..8] of integer=(-2,-1,1,2,2,1,-1,-2); dy:array[1..8] of integer=(1,2,2,1,-1,-2,-2,-1); 这个就是传说中的增量矩阵。其实也没那么神秘,就是一张表,有8种变化状态,每种状态对应了一个delta x和y...

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