A SUPER SIMPLE ADDITION PROBLEM FOR PASCAL

You're so good, readln, it's not like you can only read one data. Instead, it reads a specified amount of data, such as readln(a,b); Only two data can be read and stored in A and B. respectively

e.g. readln(a,b,c); Only three data can be read and stored in a, b, and c

Your mistake should be that the definition of the variable after var can only be ':'

Moreover, it is recommended to save the value of the sum of a and b to the variable c, and a, b, and c are all set to long integers;

That's right**. program abcs;

vara,b,c:longint;(a, b followed by colons, and variables set to long integers).

beginreadln(a，b);

c:=a+b;

writeln(c);

end.

It's too simple, readln, can only read one data. Encounter the end of carriage return.

That's right**. program abcs;

vara,b:integer;

readln(a);

readln(b);

writeln(a+b);

end.

var a,b:integer;

beginreadln(a,b);

writeln(a+b);

end.That's it.

No C, no program

Yes, the answer is 4.

You go from n=1,2,3,4,......Keep arranging, and you can find the pattern when you count it to n=7 or 8, even if it is manual, it is very fast.

The upstairs is very clear, and it is simple to figure out the law of the cycle. (1,2,3,4,5,-1) is a cycle. 16/6＝2...4, so the end result is 4.

First of all, this is a very common problem for beginners, and if you use the general solution to enumerate, it is easy to write the following **:

If a rooster costs 5 money, there will be a maximum of 19 roosters; If a hen costs 3 dollars, there will be a maximum of 33 hens; Thus let i and j cycle from 1 19 and 1 33 respectively).

var i,j,k:integer;

beginfor i:=1 to 19 do

for j:=1 to 33 do

begink:=100-i-j;k is the number of chicks;

if k mod 3<>0 then continue;

if 5*i+3*j+k div 3=100 then

writeln(i:5,j:5,k:5);

end;end.

But if you think about it, you can do even better!

Note that since x, y, and z are all integers, I also take the whole number of the above range of x, which itself has fractions.

**Write it out like this:

var x:integer;

beginfor x:=26 to 28 do

writeln(4*x-100:5,200-7*x:5,3*x:5);

end.