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ABR144A - Фибаноччийн дараалал

Фибоначчийн дараалал u0, u1, ..., un дараах дүрмээр зохиогдоно: u0=0; u1=1; ui=ui-1+ui-2 (i=2, 3, ...). n>1 натурал тоо өгөгдөв. u0, u1, ..., un – г ол.

Input

n>1 натурал тоо өгөгдөнө

Output

u0, u1, ..., un – г тус бүрийг нь нэг мөрөнд доош нь цувуулан хэвлэнэ

Example

Input:
5

Output:
0
1
1
2
3
5

Нэмсэн:sw40
Огноо:2007-12-31
Хугацааны хязгаарлалт:1s
Эх кодын хэмжээний хязгаарлалт:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Програмчлалын хэлүүд:Бүгд дараах хэлүүдээс бусад: ASM32-GCC MAWK BC C-CLANG NCSHARP CPP14 CPP14-CLANG COBOL COFFEE D-CLANG D-DMD DART ELIXIR ERL FANTOM FORTH GOSU GRV JS-RHINO JS-MONKEY JULIA KTLN NIM NODEJS OBJC OBJC-CLANG OCT PERL6 PICO PROLOG PYPY PYPY3 PY_NBC R RACKET RUST CHICKEN SQLITE SWIFT UNLAMBDA VB.NET
Эх сурвалж:Абрамов С. А.

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2024-02-16 04:09:21
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Last edit: 2024-02-16 06:21:49
2024-02-16 04:02:59


Last edit: 2024-02-16 04:09:47
2019-05-26 22:54:22
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();

int[] a = new int[n+2];

a[0] = 0;
a[1] = 1;
System.out.println(a[0] + "\n" + a[1]);
for(int i=2; i<=n; i++){
if(n > 1){
a[i] = a[i-1] + a[i-2];
}

System.out.println(a[i]);
2018-08-22 05:48:22
#include <cstdio>

int main() {
int b,i,n,a[100];
scanf("%d",&n);
printf("0");
printf("\n");
printf("1");
a[0] = 0;
a[1] = 1;
printf("\n");
for(i = 2; i <= n; i++){
a[i]=a[i-1]+a[i-2];
printf("%d",a[i]);
printf("\n");
}

return 0;
}
2018-03-01 03:47:07
medeej nemegdej yawnashdee
2011-01-12 04:26:08 sw10d019
hm.. u[i]=u[i-1]+u[i-2] gej bgaa ym u? esvel u[i]=i[u]-1+u[i-2] gej bgaa ym u???
2010-10-03 14:05:51 Eegiism.[sw09d412]
Bodoltiig ni barag bichtsen bnashdee kkk
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