jva 104 河洛之數

1) 矩陣內數字皆不可重複出現% q S, Z$ y: W: f3 }3 @0 k
(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 j1029 W7 j! W" Y% t6 S4 z0 ?, I1 K
{/ x) a6 Y, e7 c4 p( @
public static void main(String arg[])
{7 X' G. t/ G5 Q) u4 }( k
      int num []={1,2,3,4,5,6,7,8,9};8 d$ n- [( C* r; l7 K
      while(trun)6 i7 q0 s5 @4 \$ S% t- C7 v
{
int r=(int)(Math.random()*9)+1;
int t=num[i];
num[i]=num[r];
num[r]=t;# g, Q! o1 V+ k4 R
int a1 = num[0] + num[1] + num[2] ;& X4 F- `  n( @) W$ ^7 L8 e
      int a2 = num[3] + num[4] + num[5] ;
      int a3 = num[6] + num[7] + num[8] ;" ]3 u: s2 W9 H6 H& q
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;) S5 ^6 u9 l8 m$ j' o- C
      int a6 = num[2] + num[5] + num[8] ;3 [3 z; T' c0 x* h/ y
      int a7 = num[0] + num[4] + num[8] ;
      int a8 = num[2] + num[4] + num[6] ;+ _$ i, i7 ]9 r
if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)
      }break;
}% u. D+ ^1 z7 d- s/ n/ D' _. V5 r
}

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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+ s2 \( b1 f/ V- z3 j* k
{
public static void main(String arg[])
      {
         int num []={1,2,3,4,5,6,7,8,9};
      while(trun)
            {
            int r=(int)(Math.random()*9)+1;# T3 _! B+ D. u
            int t=num[i];3 M/ ?7 M1 u0 Q5 G
            num[i]=num[r];  ^- l; ]! z( R3 J0 o" P
            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] ;
      int a5 = num[1] + num[4] + num[7] ;
      int a6 = num[2] + num[5] + num[8] ;: u$ R; X" N5 K( s, t3 c! m- s
      int a7 = num[0] + num[4] + num[8] ;( V. }; }% |5 w
      int a8 = num[2] + num[4] + num[6] ;5 w, ?- [: B. o3 S  }. Q8 }
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)
         }break;          8 _  S* A. v2 z8 ?
}
}

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