type
ar = array of array of integer;
var
sum:array [1..6] of integer;
n,i,j:integer;
a: ar;
begin
write('N: '); readln(n);
setlength(a,n,n);
dec(n);
for i:=0 to n do
for j:=0 to n do
begin
readln(a[i,j]);
if i+j<n then sum[5]:=sum[5]+a[i,j]
else
if i+j>n then sum[6]:=sum[6]+a[i,j]
else sum[2]:=sum[2]+a[i,j];
if i<j then sum[3]:=sum[3]+a[i,j]
else
if j<i then sum[4]:=sum[4]+a[i,j]
else
sum[1]:=sum[1]+a[i,j];
end;
writeln;
for i:=0 to n do
begin
for j:=0 to n do
write(a[i,j]:4);
writeln;
end;
writeln;
for i:=1 to 6 do
write(sum[i]:6);
end.
Если решение оказалось полезным, пометьте его лучшим
#include <iostream>
#include <Windows.h>
int main()
{
int v;
double S, t, t1, tmpT,tmpT1,b,tmpT3;
std::cout << "V - km\h: ";
std::cin >> v;
std::cout << "\nS - km: ";
std::cin >> S;
std::cout << "\nT - hour: ";
std::cin >> t;
std::cout << "\nT1 - minut: ";
std::cin >> t1;
tmpT = S / v;
t -= tmpT;
tmpT3 = t - tmpT;
if (tmpT3 < t)
{
std::cout << "edem bez ostanovok" << std::endl;
system("pause");
exit(1);
}
tmpT1 = (t * 60) / t1;
b = (double)(int)tmpT1;
std::cout << "kol ostanovok: " << b << std::endl;
system("pause");
}