text.cpp

利用矩阵的思想 大量数据分类储存 以矩阵的形式储存 大家可以参考下

#include <iostream>
#include <fstream>
#include <stdlib.h>
#include <time.h>
using namespace std;
int main(){
   int a1=0,a2=0,a3=0,b1=0,b2=0,b3=0,j=0,i=0,c1=0,c2=0,c3=0,k=0;
   static int A[110000][3],B[110000][3];
   for (a1=0;a1<7;a1++)
   {
	for (a2=0;a2<7;a2++)
	{ 
	 for (a3=0;a3<7;a3++)
	 {
	  for (b1=0;b1<7;b1++)
	  {
	    for (b2=0;b2<7;b2++)
	    {
		  for (b3=0;b3<7;b3++)
		  {
		   if (((a2*b1-a2*b3-a3*b1-a1*b2+a1*b3+a3*b2)%7!=0)&&(a1!=a3)&&(a1!=a2)&&(a2!=a3)&&(b1!=b2)&&(b1!=b3)&&(b2!=b3))
		   {   //cout<<a1<<" "<<a2<<" "<<a3<<"      "<<b1<<"  "<<b2<<"  "<<b3<<endl;
		      /*cout<<" 1 "<<a1<<" "<<b1<<endl;
              cout<<" 1 "<<a2<<" "<<b2<<endl;
			  cout<<" 1 "<<a3<<" "<<b3<<endl;
			  cout<<"***********************"<<endl;*/
			  A[i][0]=1;
			  A[i][1]=a1;
			  A[i][2]=b1;
			  i++;
              A[i][0]=1;
			  A[i][1]=a2;
			  A[i][2]=b2;
			  i++;
              A[i][0]=1;
			  A[i][1]=a3;
			  A[i][2]=b3;
			  i++;			  
			  //out<<a1<<" "<<a2<<" "<<a3<<"       "<<b1<<"  "<<b2<<"  "<<b3<<endl;  
		   }
		  }
	    }
	  }
	 }
	}
   }  
  // srand(time(NULL));
   //          do {c1=rand()%i;
//			    c2=rand()%i;
	//		    c3=rand()%i;
             	
    //         } while(c1==c2||c2==c3||c1==c3);
//			 if((A[c2][1]*A[c1][2]-A[c2][1]*A[c3][2]-A[c3][1]*A[c1][2]-A[c1][1]*A[c2][2]+A[c1][1]*A[c3][2]+A[c3][1]*A[c2][2])!=0)
//			 {
              /* ofstream out("a.txt",fstream::out|fstream::app);
              for(j=0;j<105840;j++)
			  {
              out<<" 1 "<<A[j][1]<<" "<<A[j][2]<<endl;
			  cout<<" 1 "<<A[j][1]<<" "<<A[j][2]<<endl;//将所有结果输出置txt文档和dos界面
			  }
			  cout<<"结果共有 "<<i<<" 行";*/	  
              B[0][0]=A[0][0];
              B[0][1]=A[0][1];
			  B[0][2]=A[0][2];
			  B[1][0]=A[1][0];
              B[1][1]=A[1][1];
			  B[1][2]=A[1][2];
			  B[2][0]=A[2][0];
              B[2][1]=A[2][1];
			  B[2][2]=A[2][2];
			  cout<<B[0][0]<<B[0][1]<<B[0][2]<<endl;
			  cout<<B[1][0]<<B[1][1]<<B[1][2]<<endl;
			  cout<<B[2][0]<<B[2][1]<<B[2][2]<<endl;//由A向B复制前三行
			  do {
				  for (k=4;k<105840;k++)
                  for (i=0;i<k;i++)
				  {j=i+1;

				  } 
			  } 
return 0;
}