思路：這遞迴遞的我頭都暈了，先貼個偽遞迴吧

AC程式碼：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

void fun1(int a) {		//輸出2的a次方，因為題目最大就20000,2的15次方足以 
	if(a == 0) printf("2(0)");
	else if(a == 1) printf("2");
	else if(a == 2) printf("2(2)");
	else if(a == 3) printf("2(2+2(0))");
	else if(a == 4) printf("2(2(2))");
	else if(a == 5) printf("2(2(2)+2(0))");
	else if(a == 6) printf("2(2(2)+2)");
	else if(a == 7) printf("2(2(2)+2+2(0))");
	else if(a == 8) printf("2(2(2+2(0)))");
	else if(a == 9) printf("2(2(2+2(0))+2(0))");
	else if(a == 10) printf("2(2(2+2(0))+2)");
	else if(a == 11) printf("2(2(2+2(0))+2+2(0))");
	else if(a == 12) printf("2(2(2+2(0))+2(2))");
	else if(a == 13) printf("2(2(2+2(0))+2(2)+2(0))");
	else if(a == 14) printf("2(2(2+2(0))+2(2)+2)");
	else if(a == 15) printf("2(2(2+2(0))+2(2)+2+2(0))");
}

void fun2(int a, int cur) {
	if(a == 0) return;
	int t = a % 2;
	int tmp = a / 2;
	fun2(tmp, cur + 1);
	if(t == 1) {
		if(tmp != 0) printf("+");
		fun1(cur);
	}
}

int main() {
	int n;
	while(scanf("%d", &n) != EOF) {
		fun2(n, 0);
		printf("\n");
	}
	return 0;
}