jva 104 河洛之數

1) 矩陣內數字皆不可重複出現' r" M, v# d& a6 `, ?7 R
(2) 程式每次執行，矩陣數字排列皆不相同% p/ ?! V3 J' h! m
(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[])
{5 B' E. L1 t7 q& n
      int num []={1,2,3,4,5,6,7,8,9};
      while(trun)
{
int r=(int)(Math.random()*9)+1;1 G( z9 _7 B' H
int t=num[i];
num[i]=num[r];, d2 v; O! Y; A4 k% k" V2 @* A* L
num[r]=t;1 O, b+ T3 k# v
int a1 = num[0] + num[1] + num[2] ;
      int a2 = num[3] + num[4] + num[5] ;6 ~8 R# T! F. r: K. Z7 M2 b- Q2 l% F/ \
      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] ;! l- [8 F6 j9 R' d/ O
      int a7 = num[0] + num[4] + num[8] ;) Z* V/ ^. [( m  A) P1 u1 g1 n
      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;       1 |$ a; Z/ Z) P. a) p, V
}
}

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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" J; f( C- Q& }' ^2 W+ o
{3 N+ ~2 J0 \  d. a
public static void main(String arg[])( p) G  B& O4 O  S- ?2 T
      {
         int num []={1,2,3,4,5,6,7,8,9};- e( |! N( C8 ^% M
      while(trun)5 [* H& |0 j2 E7 S/ D- f" N
            {) M/ Z& p& e: @( ?; |4 }! A
            int r=(int)(Math.random()*9)+1;3 v4 h. g; B. I3 Y1 z* y. K8 h
            int t=num[i];9 q2 ^2 ]+ P- G
            num[i]=num[r];
            num[r]=t;
            int a1 = num[0] + num[1] + num[2] ;
      int a2 = num[3] + num[4] + num[5] ;9 Z8 X4 ]$ N: z: S$ N- P) i3 Z
      int a3 = num[6] + num[7] + num[8] ;& `0 k  W; Q+ j- e
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;6 z7 z+ \* }0 {" b2 y
      int a6 = num[2] + num[5] + num[8] ;! P6 g$ n% c) r8 Q
      int a7 = num[0] + num[4] + num[8] ;
      int a8 = num[2] + num[4] + num[6] ;1 w1 E3 p* X& ^( k& Y& O
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)8 H; t. g: y( P9 E/ y' @
         }break;          6 M" x* w) b( e3 S  ~6 j
}
}

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