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
(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# J* m! S9 P& D. \2 b; j' ^
{
public static void main(String arg[])
{
      int num []={1,2,3,4,5,6,7,8,9};7 G1 x2 `" A7 m+ B: x1 a
      while(trun)
{3 ]7 ^5 }, y9 e( D9 P8 [! h( A* k
int r=(int)(Math.random()*9)+1;
int t=num[i];
num[i]=num[r];/ Z5 d  U9 O8 s1 p; E* d
num[r]=t;  a# P0 R( T2 S) a- r' i4 Y; b
int a1 = num[0] + num[1] + num[2] ;/ [' J5 K; s4 L
      int a2 = num[3] + num[4] + num[5] ;3 R8 v6 l- I9 \4 b5 }: g
      int a3 = num[6] + num[7] + num[8] ;0 |& m& |$ x. z9 @3 n. h& s
      int a4 = num[0] + num[3] + num[6] ;8 R  e6 g0 U9 w
      int a5 = num[1] + num[4] + num[7] ;$ V* A0 o/ {' i$ m2 q
      int a6 = num[2] + num[5] + num[8] ;# W. ^4 L& ?1 ^6 b
      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), j) `" A6 p4 h0 k# ]
      }break;
}  d0 H9 y' H; a2 M& `/ R; a5 {: I1 ~
}

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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. J$ B# _& ^$ u# \; O5 W( z; K7 @
{
public static void main(String arg[])
      {
         int num []={1,2,3,4,5,6,7,8,9};
      while(trun)2 h$ k9 G% t# f! d# @! B
            {
            int r=(int)(Math.random()*9)+1;
            int t=num[i];
            num[i]=num[r];+ p8 q* l/ J% }  F3 c
            num[r]=t;. k0 J, b- u; g) V0 X. R
            int a1 = num[0] + num[1] + num[2] ;/ a: Z* `+ |+ p; Z; w1 v4 L0 @
      int a2 = num[3] + num[4] + num[5] ;
      int a3 = num[6] + num[7] + num[8] ;
      int a4 = num[0] + num[3] + num[6] ;/ J1 i$ A$ a1 j  y
      int a5 = num[1] + num[4] + num[7] ;/ ?4 p9 U. Z8 `$ q! Q, H
      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)! R8 l- c: x! s5 ~3 ]
         }break;          9 v$ v) D$ a" L& ~  K
}, \: A% i$ s- V9 \8 p
}

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