Pascal is programmed to input n and m, output n to the m power, and the answer will not exceed the l

varn,m,i:integer;

t:int64;

s:longint;

x:boolean

beginx:=false;

while x=false do

beginwrite('bottom number (n):');

readln(n);

write('exponent (m):');

readln(m);

t:=1;for i:=1 to m do t:=t*n;Powered by.

if t > 2147483648 then to determine whether the result is outside the longint range.

beginx:=false;

writeln('error!');

endelse begin

x:=true;

s:=t;write(n,'^',m,'=',s);

end;end.

First of all, the landlords trust me to have a program that is shorter and better than theirs:

1:program exam;var i,x:integer;beginx:

1;for i:=1 to 99 dox:=x*99;writeln(x mod 1000);:

This question is more classic, and my method is the simplest in history, referred to as the sieve method.

program zhishu;

var x,n: longint;

var j:boolean;

beginwhile not eof do

beginreadln(n);

if (n=0) or (n=1) then writeln('n','o') else

beginx:=2;

j:=true;

for x:=2 to round(sqrt(n)) doif n mod x = 0 then j:=false;

if j=true then writeln('yes') else writeln ('no');

end;end;

end.The watchtower lord adopted.

Troublesome. Adopt. ·.

You are chenquxin, this question is too big data, with high precision. Program: Program P1036; var i,j,k,m,n:

longint; a,b,c:array[1..25] of longint; x,y,l:

int64; begin readln(x,y); if x mod 2=1 then inc(x); if y mod 2=1 then dec(y); if x>y then writeln('0') else begin l:=(y-x) div 2+1;x:=x+y;y:

l;m:=0;n:=0; if x mod 2=0 then x:

x div 2 else y:=y div 2; repeat inc(m); a[m]:=x mod 10;x:

x div 10; until x=0; repeat inc(n); b[n]:=y mod 10;y:=y div 10; until y=0; for i:

1 to m do for j:=1 to n do c[i+j-1]:=c[i+j-1]+a[i]*b[j]; for j:

1 to n+m do begin c[j+1]:=c[j+1]+c[j] div 10; c[j]:=c[j] mod 10; end; l:

m+n; while c[l]=0 do dec(l); for i:=l downto 1 do write(c[i]);end; end.

The reason for the upstairs error is that the longint is only 10 digits, which is not enough range. You try to use real, and write write(s,0,0) when output; If the number is too large and the scientific notation is output, the only way to do this is to take some time to write high-precision.

Hello landlord.

Lucky to see your problem.

I can't help you enough!

It may be that the question you are asking is a bit professional. Or someone else hasn't encountered or touched your problem, so they can't help you. It is recommended that you go to the relevant forum for help, where there are usually more people and more enthusiastic, and may be able to help you solve the problem quickly.

Good luck. Finally, I wish you a happy, healthy and happy family every day!

Pure simulation on the line,2013julyxx hit the program should be right,But what is your n range?,There is another situation that is-no matter how much m is,Can't meet the requirements,What is the output at this time?

var a:array [1..1000000] of longint;b:array [1..1000000] of boolean;

i,j,n,m,s:longint;

beginreadln(n,m);for i:=2 to m do begin a[i]:=i;b[i]:=true;end;i:=1;

repeat

i:=i+1;if b[i]=true then for j:=2 to m div i do b[j*i]:=false;

until i=m;

for i:=2 to m do if (b[i]=true)and(a[i]>=n) then s:=s+1;writeln(s);

end.May I ask Li Zhe, did you do it on Fenghua? In other words, if it is, it is right to grind potatoes.

82872 dengke 1153 accepted 5416k 61ms pascal 2009-11-06 14:09:18

You can only use the Miller Rabin prime number test (Lead Lu remembers seeing it in "The Art of Sakura Fantasy Algorithm").

According to a large number of statistics, if a (n-1)mod n=1 is satisfied, then the probability of spinal combustion is prime.

There's a C++ here, but the idea should be the same.

Does it also time out to do it with the screening method?

You can also add some methods to the push function to execute, for example:

view plaincopy

var visitor_id;

function()

After recursion bz:=true, bz may be assigned to false again in later results, so an error will occur. The modifications are as follows, if you don't understand, you can send a private message or ask.

vara:array[1..10] of longint;

b:array[1..10] of longint;

n,m,i:longint;

procedure sum(n,m:longint);

beginif m=0 then begin

writeln('yes');

halt;endelse

beginif n=1 then exit elsebegin

sum(n-1,m-a[n]);

sum(n-1,m);

end;end;

end;begin

readln(n);

for i:=1 to n do

read(a[i]);

readln(m);

sum(n,m);

writeln('no');

end.