jva 104 河洛之數

1) 矩陣內數字皆不可重複出現9 ^4 }$ l. E1 y; F5 v: H) E& \
(2) 程式每次執行，矩陣數字排列皆不相同; y! n3 d- [/ L( Q( Q# V# f
(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 j1022 R6 T' b: H7 h! i
{
public static void main(String arg[])/ N- L0 z2 L$ j
{: h. S1 r2 m, ?- m0 B+ p
      int num []={1,2,3,4,5,6,7,8,9};
      while(trun): q9 j3 B0 U. k/ t( D
{
int r=(int)(Math.random()*9)+1;% _7 m$ N6 E! F7 N
int t=num[i];
num[i]=num[r];
num[r]=t;5 n6 G) D9 K8 Q% ?
int a1 = num[0] + num[1] + num[2] ;
      int a2 = num[3] + num[4] + num[5] ;
      int a3 = num[6] + num[7] + num[8] ;% g+ C9 P: K9 D1 A6 g$ a
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;
      int a6 = num[2] + num[5] + num[8] ;9 E; ]& e6 N/ Y9 F
      int a7 = num[0] + num[4] + num[8] ;
      int a8 = num[2] + num[4] + num[6] ;
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)
      }break;       8 G6 Z/ E. @1 E
}
}

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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
{7 l# X. x7 w* C9 M
public static void main(String arg[])8 C' |* U, V4 o  L+ V
      {
         int num []={1,2,3,4,5,6,7,8,9};
      while(trun)
            {
            int r=(int)(Math.random()*9)+1;
            int t=num[i];2 O5 \- t! g! }
            num[i]=num[r];
            num[r]=t;
            int a1 = num[0] + num[1] + num[2] ;) J8 }; e' g9 F
      int a2 = num[3] + num[4] + num[5] ;0 i: G! G  g7 E. M
      int a3 = num[6] + num[7] + num[8] ;
      int a4 = num[0] + num[3] + num[6] ;  x" M4 z+ m( \8 O6 y1 s5 E
      int a5 = num[1] + num[4] + num[7] ;
      int a6 = num[2] + num[5] + num[8] ;
      int a7 = num[0] + num[4] + num[8] ;
      int a8 = num[2] + num[4] + num[6] ;
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)3 W# d* r; P$ {
         }break;
}+ d6 v8 `+ T* y; ]' m
}

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