用主函数调用函数 求二唯数阻的和 在掉用函数中完成求和

#include<iostream.h>

int Sum(int array[][3],int a,int b)
{
int sum=0;
for(int i=0;i<a;i++)
{
for(int j=0;j<b;j++)
{
sum+=array[i][j];
}
}
return sum;
}

void main()
{
int a[3][3]={{1,2,3},{1,2,3},{1,2,3}};
int b=Sum(a,3,3);
cout<<b;
}

#include <iostream>

template<class T>
T sum(T* array, int width, int height)
{
T sum = 0;
for(int i=0;i<width*height;i++)
sum+=array[i];
return sum;
}

int main()
{
int a[3][3]={{1,2,3},{1,2,3},{1,2,3}};
std::cout<<sum<int>((int*)a, 3,3)<<std::endl;
}

const
n=10;
var
i,j,s:integer;
function qiu(i,j:integer);
begin
s:=s+q[i,j];
end;
begin
s:=0;
for i:=1 to n do
for j:=1 to n do
read(q[i,j]);
for i:=1 to n do<