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 Z5 S( N" \% `5 _ v
(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
{
public static void main(String arg[])
{# _' d6 x4 F: E
      int num []={1,2,3,4,5,6,7,8,9};
      while(trun)
{# A9 X6 M& L. I7 g; Y( u: c
int r=(int)(Math.random()*9)+1;
int t=num[i];) ^& ~/ i3 p& G: S) @" M- ^/ E; W
num[i]=num[r];
num[r]=t;
int a1 = num[0] + num[1] + num[2] ;
      int a2 = num[3] + num[4] + num[5] ;
      int a3 = num[6] + num[7] + num[8] ;
      int a4 = num[0] + num[3] + num[6] ;, r. P- h! t7 i2 a/ }! H
      int a5 = num[1] + num[4] + num[7] ;
      int a6 = num[2] + num[5] + num[8] ;: }2 R" W8 l" q2 l
      int a7 = num[0] + num[4] + num[8] ;/ F0 @1 J* b4 v3 t% X% _
      int a8 = num[2] + num[4] + num[6] ;4 \. B0 V+ w1 _& t5 m7 Q
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)- B/ ~1 M8 ?7 U8 q, }
      }break;
}1 ~, ^9 [' \- c6 x1 e3 h0 i
}

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新手上路

Rank: 1

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
{4 M7 G, A4 G: [9 @1 y  ?' p0 J4 s) O
public static void main(String arg[])
      {
         int num []={1,2,3,4,5,6,7,8,9};
      while(trun)
            {3 O' c% ^5 \7 Y6 g% q/ @" w" U: S
            int r=(int)(Math.random()*9)+1;) z9 R$ }; C, W: s5 W. @! w% N
            int t=num[i];
            num[i]=num[r];
            num[r]=t;/ S8 f6 l; g6 a* Z9 X
            int a1 = num[0] + num[1] + num[2] ;! D* @, ], Q) K6 |6 A# l
      int a2 = num[3] + num[4] + num[5] ;
      int a3 = num[6] + num[7] + num[8] ;6 N6 l  y' d6 J. k; s& p
      int a4 = num[0] + num[3] + num[6] ;! {6 b  i& E3 e% l' _6 Z
      int a5 = num[1] + num[4] + num[7] ;, Z: v7 ^) V' i8 }
      int a6 = num[2] + num[5] + num[8] ;
      int a7 = num[0] + num[4] + num[8] ;
      int a8 = num[2] + num[4] + num[6] ;0 Y. x0 l  H( D# r
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)* a& {% ^3 H1 z/ J* x- ?7 V
         }break;
}3 b* }4 f( E2 O1 u  h
}

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