HOJ 6433 Problem H. Pow

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Problem H. Pow Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others) Total Submission(s): 1521    Accepted Submission(s): 721   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

知识点：string

#include<iostream>
#include<cstdio>
#include<string>

using namespace std;

string str[1010];

void init()
{
	str[0] = "1";
	bool flag = false;
	for(int i=1; i<=1000; i++) {
		string s = "";
		for(int j=str[i-1].size()-1; j>=0; j--) {
			int t = (str[i-1][j]-'0')*2;
			if(flag) {
				t++;
				flag = false;
			}
			if(t<10) {
				s = (char)(t+'0') + s;
			}else {
				s = (char)(t-10+'0') + s;
				flag = true;
			}
		}
		if(flag) s = "1" + s;
		str[i] = s;
		flag = false;
	}
}

int main()
{
	int T_T;
	init();
	cin >> T_T;
	while(T_T--) {
		int n;
		cin >> n;
		cout << str[n] <<endl;
	}
	return 0;
}