jva 104 河洛之數

1) 矩陣內數字皆不可重複出現
(2) 程式每次執行，矩陣數字排列皆不相同1 N' N5 l! k, d; a
(3) 矩陣之縱向、橫向、對角線數字總和皆為15

#include <iostream>
#include <cstdlib>
using namespace std ;
int main(){
int a[] = {1,2,3,4,5,6,7,8,9} ;
//亂數種子
srand(time(NULL));
while(true){
//交換9次
for(int i=0;i<9;i++){
int r = rand()%9 ;
//交換
int t = a[r] ;
a[r] = a[i];
a[i] = t ;
}
//列的判斷
int a1 = a[0] + a[1] + a[2] ;
int a2 = a[3] + a[4] + a[5] ;
int a3 = a[6] + a[7] + a[8] ;
//行的判斷
int a4 = a[0] + a[3] + a[6] ;
int a5 = a[1] + a[4] + a[7] ;
int a6 = a[2] + a[5] + a[8] ;
//斜的判斷
int a7 = a[0] + a[4] + a[8] ;
int a8 = a[2] + a[4] + a[6] ;
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15){
break;
}
}
cout << a[0] << " " << a[1] << " " << a[2] << endl;
cout << a[3] << " " << a[4] << " " << a[5] << endl;
cout << a[6] << " " << a[7] << " " << a[8] << endl;
system("pause");
return 0;
}

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2^#

johnson發表於 2011-12-31 11:35 | 只看該作者

public class j102
{" m* P# ?/ Y3 K" l/ Z; }1 i5 \$ N
public static void main(String arg[])7 C: L& d4 P1 g$ j" D
{
      int num []={1,2,3,4,5,6,7,8,9};
      while(trun)$ ^* V' @) ~# @& M5 h
{7 S. X, G0 t$ Q7 P. g' m( J( ?+ F  f
int r=(int)(Math.random()*9)+1;# \. L# Y% ?) ^2 y( b
int t=num[i];
num[i]=num[r];
num[r]=t;
int a1 = num[0] + num[1] + num[2] ;& W1 c! x3 J& S) C: F  J
      int a2 = num[3] + num[4] + num[5] ;
      int a3 = num[6] + num[7] + num[8] ;
      int a4 = num[0] + num[3] + num[6] ;( f% h) Q& d4 z, N2 i8 O6 _
      int a5 = num[1] + num[4] + num[7] ;
      int a6 = num[2] + num[5] + num[8] ;
      int a7 = num[0] + num[4] + num[8] ;9 Q4 y3 p+ S8 f( I; j
      int a8 = num[2] + num[4] + num[6] ;  B* O/ c8 Q& F% {
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)
      }break;
}
}

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3^#

kim發表於 2011-12-31 11:36 | 只看該作者

#include <iostream>
#include <cstdlib>
using namespace std ;
int main(){
int a[] = {1,2,3,4,5,6,7,8,9} ;
//亂數種子
srand(time(NULL));
while(true){
//交換9次
for(int i=0;i<9;i++){
int r = rand()%9 ;
//交換
int t = a[r] ;
a[r] = a[i];
a[i] = t ;
}
//列的判斷
int a1 = a[0] + a[1] + a[2] ;
int a2 = a[3] + a[4] + a[5] ;
int a3 = a[6] + a[7] + a[8] ;
//行的判斷
int a4 = a[0] + a[3] + a[6] ;
int a5 = a[1] + a[4] + a[7] ;
int a6 = a[2] + a[5] + a[8] ;
//斜的判斷
int a7 = a[0] + a[4] + a[8] ;
int a8 = a[2] + a[4] + a[6] ;
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15){
break;
}
}
cout << a[0] << " " << a[1] << " " << a[2] << endl;
cout << a[3] << " " << a[4] << " " << a[5] << endl;
cout << a[6] << " " << a[7] << " " << a[8] << endl;
system("pause");
return 0;
}

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4^#

eric5802發表於 2011-12-31 11:38 | 只看該作者

public class j102
{, a; y' n+ B: @- A5 x6 j! z
public static void main(String arg[])
      {1 y* s5 x" w: t' h
         int num []={1,2,3,4,5,6,7,8,9};1 ~' O4 T  D# n4 Z
      while(trun)7 o$ K7 J4 K* }" J' d7 ?* C
            {4 u: ~  |& ^9 g. v# o
            int r=(int)(Math.random()*9)+1;" C, r# {1 r! F9 m! O
            int t=num[i];
            num[i]=num[r];/ m0 z) `$ F* J/ c2 [
            num[r]=t;
            int a1 = num[0] + num[1] + num[2] ;; P9 z5 X( a+ T, o
      int a2 = num[3] + num[4] + num[5] ;* Y: l3 w) p9 O# g1 q
      int a3 = num[6] + num[7] + num[8] ;
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;
      int a6 = num[2] + num[5] + num[8] ;0 t/ a; B! }0 D1 Q0 [
      int a7 = num[0] + num[4] + num[8] ;% t1 W* `2 J) \7 ^
      int a8 = num[2] + num[4] + num[6] ;" z! f( ?" G5 e8 z! y4 K
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)
         }break;          / X! n- }0 Q6 I2 c% {
}
}

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