九度OJ 1095：2的幂次方 （递归）

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：913

解决：626

题目描述：

    Every positive number can be presented by the exponential form.For example, 137 = 2^7 + 2^3 + 2^0。

    Let's present a^b by the form a(b).Then 137 is presented by 2(7)+2(3)+2(0). Since 7 = 2^2 + 2 + 2^0 and 3 = 2 + 2^0 , 137 is finally presented by 2(2(2)+2 +2(0))+2(2+2(0))+2(0). 
 
    Given a positive number n,your task is to present n with the exponential form which only contains the digits 0 and 2.

输入：

    For each case, the input file contains a positive integer n (n<=20000).

输出：

    For each case, you should output the exponential form of n an a single line.Note that,there should not be any additional white spaces in the line.

样例输入：

1315

样例输出：

2(2(2+2(0))+2)+2(2(2+2(0)))+2(2(2)+2(0))+2+2(0)

来源：

2006年上海交通大学计算机研究生机试真题

思路：
需用递归来做，注意边界条件，2不应该被打印成2(0).

代码：

#include <stdio.h> void present(int n){    if (n == 0 || n == 2)    {        printf("%d", n);        return;    }    int i;    int a[20], c;    for (i=0; n>0; i++)    {        a[i] = n%2;        n /= 2;    }    c = i;    for (i=c-1; i>=0; i--)    {        if (a[i] == 0)            continue;        if (i != c-1)            printf("+");        if (i == 1)        {            printf("2");            continue;        }        printf("2(");        present(i);        printf(")");    }} int main(void){    int n;     while (scanf("%d", &n) != EOF)    {        present(n);        printf("\n");    }     return 0;}/**************************************************************    Problem: 1095    User: liangrx06    Language: C    Result: Accepted    Time:0 ms    Memory:912 kb****************************************************************/