jva 104 河洛之數

1) 矩陣內數字皆不可重複出現* c& U5 n6 }+ ~1 z
(2) 程式每次執行，矩陣數字排列皆不相同0 v! H* _* E( {% @: T( f4 p8 L
(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  O( T3 W: l/ `
{* |2 B# K% P' m6 d- e* k2 h" E9 j) L
public static void main(String arg[])
{
      int num []={1,2,3,4,5,6,7,8,9};
      while(trun): ~) L& R6 @: h2 j* ~% J
{
int r=(int)(Math.random()*9)+1;4 e0 B0 U: I! f+ T
int t=num[i];
num[i]=num[r];
num[r]=t;
int a1 = num[0] + num[1] + num[2] ;: R6 P8 `8 {1 q; x
      int a2 = num[3] + num[4] + num[5] ;
      int a3 = num[6] + num[7] + num[8] ;
      int a4 = num[0] + num[3] + num[6] ;% ?0 B, Y- q) k8 ?1 @
      int a5 = num[1] + num[4] + num[7] ;/ i8 u! N4 w2 \) |. I6 [- s
      int a6 = num[2] + num[5] + num[8] ;, v1 f7 W( J( t6 F% E( Z
      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;
}
}

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新手上路

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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& ?5 e: ?0 S, f2 N
{
public static void main(String arg[])2 ^8 P/ i% Z. u, A$ P
      {1 U# [1 x, m6 S9 m2 n/ p' R
         int num []={1,2,3,4,5,6,7,8,9};3 a' C( j$ h6 s) f
      while(trun)
            {0 [! P8 }. Q# E3 [: V
            int r=(int)(Math.random()*9)+1;
            int t=num[i];) O7 j* l; S1 M1 P# x+ M8 {
            num[i]=num[r];' `& J/ k  M2 c9 ~; t
            num[r]=t;
            int a1 = num[0] + num[1] + num[2] ;$ L! d- T) N7 W! e. V
      int a2 = num[3] + num[4] + num[5] ;
      int a3 = num[6] + num[7] + num[8] ;; Z) I% r3 F8 E/ u& P/ p+ K& i
      int a4 = num[0] + num[3] + num[6] ;; v( H& o! a" ]% K7 I& F
      int a5 = num[1] + num[4] + num[7] ;: d- v5 E3 G8 Q5 I
      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)/ \! ?! v) X- h- i
         }break;          & I& Y/ T; r8 Z2 D" e- }
}: V6 H5 l( N4 X  f) q: g
}

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