Heterovariance and Pascal procedures.

Basic characters in pascal: there are 86 a z (26), a z (26), 0 9 (10), and other characters.

Reserved words: 35 (without forward); Identifier: A combination of letters and numbers at the beginning of a letter;

Basic characters in c: there are 96 a z (26), a z (26), 0 9 (10), other characters (34);

Keywords: There are 32 of them, divided into four categories, data type; control statements; storage type; Other(sizeof); Identifier: A combination of letters, numbers, and underscores that begin with a letter or underscore. pascal 或{}

The standard identifiers in pascal (standard constant 3, standard type 5, standard file 2, standard function 17) are included in the identifiers, and c is scattered in various files except for the data type where the standard type is placed in the keyword. Therefore, there is a difference between the scope of reserved words and keywords. *c is a case-sensitive language, pascal is not.

Also, many languages (e.g. c) arrays are automatically defined from 0, while pascal can be defined from 1, and many functions that are already meaningful are different.

Didn't understand what you asked.

There is no need to use fast powers.

vari: longint;

t, k: longint;

flag: boolean;

v: array [0..1000] of longint;

beginread(k);

t := k;

flag := false;

if t >= 1000 then

beginflag := true;

t := t mod 1000;

end;k := 1;

fillchar(v, sizeof(v), 255);

for i := 0 to 10000 dobegin

if flag and (i <>0) thenbegin

if v[k] <1 then

beginwriteln(i + v[k]);

break;

end;v[k] := i;

end;k := k * t;

if k >= 1000 then

beginflag := true;

k := k mod 1000;

end;end;

end.

Outputs the values of a and b.

The output string is a+b

Output string a'+'b

There is a mistake, the compilation does not pass.

It's not a Sudoku solution yet!

Procedure** is attached hereto.

I made it up myself, and I think it's the best program for myself.

Note: When entering, each data in the same row is separated by spaces, 9 per line).

The ith character of the s string, e.g. s(string)="abcdef", then s[1]="a",s[3]=“c”

Strings can be manipulated as arrays, and for no apparent reason, the pascal language is designed for convenience.

0”..9"] is a collection of information game type slippery data, not an array.

Error: score:array[1..m,1..n] of real;

M can be more than 30,000 when the program is initialized, and more than 2 billion if it is a free pascal, and the value is not fixed. It can be 0, -5, 100000000000000000, or something else. Because m is a variable.

And the application array definitely needs an exact size, it can't be so casual. So the compiler will report an error. Imagine if the compiler didn't report errors and was able to run.

What should the computer do when m is -4? Or what if it's 201235612 such a very large number?

Change the approach. Estimate the maximum value of m,n in the problem, and then apply with the maximum value (e.g. m,n in the problem is up to 1000).

then apply for score:array[1..1000,1..1000] of real;

The size of the array must be fixed first, it must be constant, and m, n are variables, so the program can be changed to error.

constmax_m = 100;

max_n = 100;

varscore:array [1..max_m, 1..max_n] of real;..

Its Chinese should be:

You can't use variable definitions.

Please change m,n to quantitative.