（高精度）—hdu-6433-Problem H. Pow

Problem H. Pow Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others) Total Submission(s): 469    Accepted Submission(s): 187   Problem Description There are n numbers 3^0, 3^1, . . . , 3^n-1. Each time you can choose a subset of them (may be empty), and then add them up. Count how many numbers can be represented in this way.   Input The first line of the input contains an integer T , denoting the number of test cases. In each test case, there is a single integers n in one line, as described above. 1 ≤ T ≤ 20, 0 ≤ n ≤ 1000   Output For each test case, output one line contains a single integer, denoting the answer. Sample Input 4 9 7 8 233   Sample Output 512 128 256 13803492693581127574869511724554050904902217944340773110325048447598592   Source 2018 Multi-University Training Contest 10   Recommend chendu   |   We have carefully selected several similar problems for you:  6437 6436 6435 6434 6433    Statistic | Submit | Discuss | Note

题解：数字 n,有 0 ~ n-1 种，共有 2 的 n 次方的状态。

Java：

import java.math.*;
import java.util.*;
import java.text.*;
import java.io.*;
public class Main {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner in=new Scanner(System.in);
        int t=in.nextInt();
        while(t>0){
            BigDecimal a=new BigDecimal(2);
            int n=in.nextInt();
            if(n==0){
                System.out.println("1");
                t--;
                continue;
            }
            for(int i=0;i<n-1;i++){
                a=a.multiply(new BigDecimal(2));
            }
            System.out.println(a);
            t--;
        }
    }
}

c++:

#include<stdio.h>
#include<queue>
#include<string.h>
#include<algorithm>
#include<set>
#include<map>
#include<math.h>
#include<vector>
#include<bitset>
#include<iostream>
#include <bits/stdc++.h>
#define ullmax 1844674407370955161
#define llmax 9223372036854775807
#define intmax 2147483647
#define re register
using namespace std;
int main(){
    int t;
    cin>>t;
    while(t--){
        double n;
        cin>>n;
        cout <<fixed<<std::setprecision(0)<<pow(2,n)<<endl;
    }
    return 0;
}