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f1(n) = 2n2, f2(n) = en+1, f3(n) = √n, f4(n)=2n, f5(n) = nlog2(3), f6(n)=3log2(n)
f7(n) = nln (n), f8(n) = n√n, f9(n) = ln (n2+ 1), f10(n) = ln (√n)
B
for i:=1 to ndo begin
for j:=1 to ido A ;
end {for} ;
x A[1..n]
n>1
Algo(A,x)
trouve := faux
i:= low(A)
tant que (non trouve) et (i≤high(A)) faire
si A[i] = xalors
trouve := vrai
sinon
i:= i+ 1
fin si
fin tant que
retourner trouve
A[i] = x
n = high(A)-low(A)+1 T(n)
x
A
x
A
1
p > 1x
[1, p]N
T(n) = p−(p−1)n
pn−1
n[1, p]pn
pnn−1x
x
k∈[1, n −1] k−1
x k−x n −k[1, p]
(p−1)k−1·pn−k
(p−1)k−1x k −1pn−k
n−k[1, p]
k+ 1
k
k−1
x
x6∈ T[1..k −1]
n1
 -

n
p(p−1)n−1+
n−1
X
k=1
(p−1)k−1pn−k=pn
T(n) = 1
pn n·p(p−1)n−1+
n−1
X
k=1
k·(p−1)k−1pn−k!
S(n) =
n−1
X
k=1
k·(p−1)k−1pn−k=pn−1
n−1
X
k=1
k·p−1
pk−1
n
n−1
X
k=1
kxk−1=(n−1)xn−nxn−1+ 1
(x−1)2
S(n) = p(n−1)(p−1)n−np(p−1)n−1+pn
T(n) = p−(p−1)n
pn−1
1000 0
20 20
p= 21 n= 1000
21
;-)
3
trouve := faux
pour ivariant de low(A) à high(A)faire
si A[i] = xalors
trouve := vrai
fin si
fin pour
retourner trouve
function f(n : CARDINAL) : CARDINAL;
var
i, res : CARDINAL;
begin
if n=0 then
res := 1
else begin
res := 0;
for i := 0 to n-1 do
res := res + (i+1)*f(i);
end {if};
f := res;
end {f};

f(1) f(2) f(3) f
f(3)
unf(n) un
ukk < n n ≥1
c(n)un=f(n)
c(n) = n+
n−1
X
k=0
c(k),
n≥0
n c(n)
c(n+ 1) −c(n)n≥0
c(n+ 1) = 2c(n)+1.
c(n)n→+∞
un+1 −un
un+1 = (n+ 2)un
n≥2
un
c0(n)un
c(n)