题解 | #2的幂次方#

#include <cctype>
#include <cstdio>
#include <iostream>

using namespace std;


void func1(int n) {
    if (n == 1) {
        cout << "2(0)";
        return;
    } else {
        // 当n大于2时
        int i, total = 0;
        cout << "2";
	   // 获取次方数
        for (i = 2; i <= n; i *= 2 ) {
            total++;
        }
        if (total != 1) {
            // 将次方数递归
            cout << "(";
            func1(total);
            cout << ")";
        }
        // n - 2^total
        int number_y = n - i / 2;
        if (number_y != 0) {
            cout << "+";
            func1(n = n - i / 2);
        }
        return;

    }
}
int main() {
    int number;
    while (scanf("%d", &number) != EOF) {
        func1(number);
    }
    return 0;
}
// 64 位输出请用 printf("%lld")