1 条题解
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0
C :
#include <stdio.h> unsigned long long compute(int n) { unsigned long long sum = 0; if (n == 1) return 1; else{ sum += 2 * compute(n - 1); } return sum + 1; } int main(void) { int num = 0; while (scanf("%d", &num) == 1) { printf("%llu\n", compute(num)); } }
C++ :
#include <iostream> using namespace std; int main(){ unsigned long long a[64]; a[0] = 1; for(int i=1;i<64;i++) a[i] = 2*a[i-1] + 1; for(int n;cin>>n;) cout<<a[n-1]<<endl;}
Pascal :
//a[n]=a[n-1]*2+1 program p21144(Input,Output); const jz=10; var i,j,k,t,lena,n:integer; a:array[1..1000] of integer; procedure chen(); var i,x:integer; begin x:=0; for i:= 1 to lena do begin a[i]:=a[i]*2+x; x:=a[i] div jz; a[i]:=a[i] mod jz; end; if x>0 then begin a[i+1]:=x; inc(lena); end; end; procedure jia(); var i,j,x:integer; begin a[1]:=a[1]+1; x:=a[1] div jz; a[1]:=a[1] mod jz; i:=1; while x<>0 do begin inc(i); a[i]:=a[i]+x; x:=a[i] div jz; a[i]:=a[i] mod jz; end; if (i=lena) then lena:=i; end; begin while not eof(Input) do begin readln(n); a[1]:=1; lena:=1; for k:=2 to n do begin chen();//*2 //writeln('c',a[3],a[2],a[1],' ',lena); jia(); //+1 end; for i:=lena downto 1 do write(a[i]); writeln; end; end.
Java :
import java.util.Scanner; import java.math.*; public class Main{ public static void main(String[] args) { Scanner in=new Scanner(System.in); while(in.hasNext()) { int b = in.nextInt(); BigInteger one=new BigInteger("1"); BigInteger two=new BigInteger("2"); BigInteger x=two.pow(b).subtract(one); System.out.println(x); } }}
- 1
信息
- ID
- 2123
- 时间
- 1000ms
- 内存
- 64MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者