返回以二維數組周邊元素之和爲函數值

出二維數組周邊元素之和作爲函數值返回

第一個for循環用於計算矩陣最上一行和最下一行的總和，第二個for循環則計算除兩頭元素以外的最左一列和最右一列的元素的和。最後，將這兩個求和結果相加得到周邊元素的和。

〔源程序〕

#include

#include

#define M 4

#define N 5

int fun(int a[M][N])

{int sum=0,i;

for(i=0;i

sum+=a[0][i]+a[M-1][i];

for(i=1;i

sum+=a[i][0]+a[i][N-1];

return sum ;

}

main()

{ int aa[M][N]={{1,3,5,7,9},

{2,9,9,9,4},

{6,9,9,9,8},

{1,3,5,7,0}};

int i,j,y;

clrscr();

printf(＂The original data is :\n＂);

for(i=0;i

{for(j=0;j

printf(＂\n＂);

}

y=fun(aa);

printf(＂\nThe sum: %d\n＂,y);

printf(＂\n＂);

}

二維數組靠邊元素之和

private function Anyl(A()() AS INTEGER,byval m as integer,byval n as integer) as long

tol=sum_side(A,m,n)

print tol

if m=n then

sd=Diag(A,m)

sid=InDiag(A,m)

print sd

print sid

else

print ＂m≠n＂

end if

end function

private function Sum_side(A()() AS INTEGER,byval m_side as integer,byval n_side as integer) as long

dim sum as long

sum=0

n_side=n_side-1

for i=0 to n step 1

sum=sum+A(0)(i)

next i

m=m-2

for i=1 to m step 1

sum=sum+A(i)(0)

sum=sum+A(i)(n)

next i

m=m+1

for i=0 to n step 1

sum=sum+A(m)(i)

next i

sum_side=sum

end function

private function Diag(A()() AS INTEGER,byval m as integer) as long

dim Sum_Diag as long

m=m-1

Sum_Diag=0

FOR i=0 to m step 1

Sum_Diag=Sum_Diag+A(i)(i)

next i

Diag=Sum_Diag

end function

private function InDiag(A()() AS INTEGER,byval m as integer) as long

dim Sum_InDiag as long

m=m-1

Sum_Diag=0

FOR i=0 to m step 1

Sum_InDiag=Sum_InDiag+A(i)(m-i)

next i

InDiag=Sum_InDiag

end function

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