Minimum product (basic type)

1#include <stdio.h>2 3 intMain ()4 {5     inti,j,k;6     intT,n;//T: number of data groups N: The number of rows per row in each set of data7     inta[2][8];//two rows of data in each set of data that is used to store input8     intTemp//Intermediate temporary variable in the process for sorting each row of data9     intSum//used to record the sum of the product of two rows of dataTen      Onescanf"%d", &t);//number of input data groups A      -      for(i=t;i>0; i--)//grouping input and processing -     { thesum=0;//Clear 0 Product and -           -scanf"%d", &n);//Enter the number of rows per row in two rows of data -          +          for(j=0;j<2; j + +)//enter two rows of data -              for(k=0; k<n;k++) +scanf"%d",&a[j][k]); A  at         /*********************************************************************** - * * To achieve the minimum product, you only need to find the two rows of data not multiplied by the maximum and the minimum value multiplication, - * * Then add, so the two rows of data are sorted in descending and ascending order -         ***********************************************************************/ -          for(j=0; j<n-1; j + +)//Sort Two rows of data in descending order and ascending order by using bubble Sort method -              for(k=j+1; k<n;k++) in             { -                 if(a[0][k]>a[0][J])//Descending to                 { +temp=a[0][j]; -a[0][j]=a[0][k]; thea[0][k]=temp; *                 } $                 if(a[1][k]<a[1][J])//AscendingPanax Notoginseng                 { -temp=a[1][j]; thea[1][j]=a[1][k]; +a[1][k]=temp; A                 } the             } +              -          for(j=0; j<n;j++)//multiply the elements of the same subscript in two rows of data, and then add and Sum $sum+=a[0][j]*a[1][j]; $  -printf"%d\n", sum);//output minimum product and -     } the      -     return 0;Wuyi}

Minimum product (basic type)