1 solutions
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0
C :
#include<stdio.h> int main() { int t,n,i; double a[30]; a[0]=1; a[1]=3; for(i=2;i<30;i++) a[i]=a[i-1]+2*a[i-2]; scanf("%d",&t); while(t--) { scanf("%d",&n); printf("%.lf\n",a[n-1]); } return 0; }
C++ :
#include<stdio.h> int main() { int t,n,i; double a[30]; a[0]=1; a[1]=3; for(i=2;i<30;i++) a[i]=a[i-1]+2*a[i-2]; scanf("%d",&t); while(t--) { scanf("%d",&n); printf("%.lf\n",a[n-1]); } return 0; }
Pascal :
program p1059; var t,i,j,k,n:longint; a:array[1..30] of longint; begin readln(t); a[1]:=1; a[2]:=3; for i:=3 to 30 do a[i]:=a[i-1]+2*a[i-2]; for i:=1 to t do begin readln(n); writeln(a[n]); end; end.
Java :
import java.util.*; public class Main { public static void main(String[] args){ Scanner sc=new Scanner(System.in); int a=sc.nextInt(); int[] b=new int[31]; b[1]=1; b[2]=3; for(int i=3;i<31;i++){ b[i]=b[i-1]+2*b[i-2]; } while(a--!=0){ System.out.println(b[sc.nextInt()]); } } }
Python :
# coding=utf-8 def cg(x,y): r=1 for c in range(x): r=r*(y-c)/(c+1) return int(r) a=int(input()) for i in range(a): count=0 b=int(input()) for j in range(int(b/2+1)): count=count+cg(j,b-j)*(2**j) print(count)
- 1
Information
- ID
- 657
- Time
- 1000ms
- Memory
- 32MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By