Two simple pascal questions

I just did this topic 77 is a classic Josephian problem that is very famous in algorithm design, the monkey chooses the king, the number is reported with a password, and the fox chases the rabbit is its variation.

The monkey elects the king:

program y5_6;

varn,m,s,f,t:integer;

a:array[1..100] of boolean;

beginreadln(n);

m:=3;for t:= 1 to n doa[t]:=false;

f:=0; t:=0; s:=0;

repeat

t:=t+1;

if t=n+1 then t:=1;

if a[t]=false then s:=s+1;

if s=m then

begins:=0;

a[t]:=true;

f:=f+1;

end;until

f=n-2;

for t:=1 to n do

if not(a[t]) then

beginwrite(t);

exit;end;

end.The fox chases the rabbit:

program y5_7;

var i,j:integer;

a:array[0..9] of integer;

beginfor i:= 0 to 9 do a[i]:=1;

j:=0;for i:= 1 to 1000 do beginj:=(j+i) mod 10 ;

a[j]:=0;

end;for i:= 1 to 9 do if a[i]=1 then write(i:3);

if a[0]=1 then write(' 10');

end.

1 is the first of course and he always counts 1

For example, at the beginning, 10,000 yuan is a monthly interest.

Then after the first month, due to the monthly interest, you owe the bank a total of 10,000 * (1 + yuan), and then you pay the yuan and still owe the yuan.

2nd month: due to the monthly interest arrears.

If you pay the yuan again, you still owe the yuan.

And so on. By the tenth month, the arrears of about $ are almost 0, so the monthly payment is.

My thoughts: do it with dichotomies.

Left Boundaryl Right BoundaryrMiddlem=(L+R) 2If m satisfies the condition, then L=M, otherwise R=M until the gap between l and r is within the border, if you don't ask me directly.

The number of months means that you have to repay in a few months, and the number of months * monthly interest is the interest you have to pay, plus the amount of the loan is 10,000; Divide by the number of months, and it is the monthly payment;

If you want to **, just tell me your email and I'll send it to you.