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 j1026 V, \; A6 A3 G
{- ~" T8 ?# s) b8 {. N
public static void main(String arg[])% P2 H: w4 W" k  T7 [
{2 G5 T. u9 Q+ X+ V. m
      int num []={1,2,3,4,5,6,7,8,9};
      while(trun)
{) B8 x: L: g* l: Y
int r=(int)(Math.random()*9)+1;
int t=num[i];
num[i]=num[r];# U' W7 u/ a8 W, i# g& z
num[r]=t;& E, q  ~% m+ N( x  t' |
int a1 = num[0] + num[1] + num[2] ;
      int a2 = num[3] + num[4] + num[5] ;" H: y4 y. _2 r! V1 I8 _
      int a3 = num[6] + num[7] + num[8] ;) Z8 L' s5 @$ G4 P. H/ l  p4 w7 a5 o! o4 V
      int a4 = num[0] + num[3] + num[6] ;
      int a5 = num[1] + num[4] + num[7] ;
      int a6 = num[2] + num[5] + num[8] ;& q' T* u$ s8 O1 }. G/ n+ B$ U
      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)2 z: }5 `9 j* n2 r
      }break;       & k' c6 Z/ J% `
}
}

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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
{( A/ x& I- f) A+ U( l
public static void main(String arg[])% X+ E$ k8 S: x0 y9 C, O( u
      {2 i& \+ S: |* P9 X$ B4 t9 s
         int num []={1,2,3,4,5,6,7,8,9};# n) E" t& D( n/ g
      while(trun)
            {
            int r=(int)(Math.random()*9)+1;
            int t=num[i];
            num[i]=num[r];
            num[r]=t;
            int a1 = num[0] + num[1] + num[2] ;! I: y# w* m' k% I4 V* r
      int a2 = num[3] + num[4] + num[5] ;
      int a3 = num[6] + num[7] + num[8] ;) Q7 J, r5 }5 K
      int a4 = num[0] + num[3] + num[6] ;  n5 K: o1 j. F; T0 u
      int a5 = num[1] + num[4] + num[7] ;
      int a6 = num[2] + num[5] + num[8] ;1 x$ U, x/ }9 Z0 J( Y0 S
      int a7 = num[0] + num[4] + num[8] ;* j4 A/ y( g2 M  w1 X5 m3 f
      int a8 = num[2] + num[4] + num[6] ;# I5 U3 e/ U& l& @, {3 y3 D
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)' d+ Q: A  g  c: g) a: Y  i
         }break;
}
}

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