（高精度）—hdu-6433-Problem H. Pow

最新推荐文章于 2020-03-08 19:50:52 发布

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最新推荐文章于 2020-03-08 19:50:52 发布

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本文介绍了一道计数问题的解决方法，通过选择不同数量的3的幂次和来生成不同的数字组合，利用数学和编程手段高效计算所有可能的组合数目。

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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

233

Sample Output

512

128

256

13803492693581127574869511724554050904902217944340773110325048447598592

Source

2018 Multi-University Training Contest 10

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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;
}