jva 104 河洛之數

1) 矩陣內數字皆不可重複出現2 Z0 y0 n0 H& E/ A
(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
{
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;% t1 M. M2 [" b7 U
int t=num[i];
num[i]=num[r];
num[r]=t;% X1 a) H( L  G! `/ `
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] ;+ r  u+ b, c" c" N% g' C$ k3 a
      int a5 = num[1] + num[4] + num[7] ;. C* ~# X1 e- [* j( B( G2 B
      int a6 = num[2] + num[5] + num[8] ;6 e: `* A9 ]! g8 T5 Q, 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)
      }break;       4 l' O0 A6 }) L9 z* Y
}
}

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新手上路

Rank: 1

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 A5 ]) X( S
public static void main(String arg[])5 j8 J: q8 b$ [8 |1 Y. b2 p* {, p
      {  V) Q. {" p' a+ z
         int num []={1,2,3,4,5,6,7,8,9};) R8 [+ H7 y+ ]  r' z& S. y1 c
      while(trun)
            {
            int r=(int)(Math.random()*9)+1;* f$ T+ C1 m$ ]  W& ~
            int t=num[i];! }2 P7 b  S+ l2 N0 @
            num[i]=num[r];
            num[r]=t;5 `! F& ~8 r2 \! {4 Z
            int a1 = num[0] + num[1] + num[2] ;: m4 N- c! Q: m5 V
      int a2 = num[3] + num[4] + num[5] ;2 C  j# M4 X; C4 w) k; r" @6 Z8 j
      int a3 = num[6] + num[7] + num[8] ;
      int a4 = num[0] + num[3] + num[6] ;4 a0 N! X# x' n8 P7 o  ?
      int a5 = num[1] + num[4] + num[7] ;
      int a6 = num[2] + num[5] + num[8] ;& Z7 i/ J: D9 U
      int a7 = num[0] + num[4] + num[8] ;5 v& F! W. |4 k# n2 M- F
      int a8 = num[2] + num[4] + num[6] ;  O& C- Q, \4 i  s
            if(a1==15&&a2==15&&a3==15&&a4==15&&a5==15&&a6==15&&a7==15&&a8==15)
         }break;
}
}

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