题解 | #2的幂次方#

//土尔逊Torson 编写于2023/06/02
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <vector>

using namespace std;

//次方计算 base 的 n 次方
int MyPow(int base, int n) {
	int result = 1;
	for (int i = 0; i < n; ++i) {
		result *= base;
	}
	return result;
}

// 计算把N分解成2的i幂次方数之和的系数集合，例如9=2^3+2^0,返回值就是3和0
vector<int> ToArray(int n) {
	vector<int> b;
	int i = 0;
	while (n > 0) {
		while (n - MyPow(2, i) > 0) {
			i++;
		}
		if (n < MyPow(2, i)) {
			i--;
		}
		n = n - MyPow(2, i);
		b.push_back(i);
		i = 0;
	}
	return b;
}

//构建答案
string Build(int n) {
	vector<int> a;
	a = ToArray(n);
	int size = a.size();
	string str = "";
	for (int i = 0; i < size; ++i) {
		if (a[i] == 1) {
			str += "2";
		}
		else if (a[i] == 0) {
			str += "2(0)";
		}
		else {
			str += "2(" + Build(a[i]) + ")";
		}

		if (i != size - 1) {
			str += "+";
		}
	}
	return str;
}

int main() {
	int n;
	while (scanf("%d", &n) != EOF) {
		printf("%s", Build(n).c_str());
	}
	system("pause");
	return EXIT_SUCCESS;
}
// 64 位输出请用 printf("%lld")