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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.For each case, the input file contains a positive integer n (n<=20000).
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.
1315
2(2(2+2(0))+2)+2(2(2+2(0)))+2(2(2)+2(0))+2+2(0)
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：
#includevoid 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****************************************************************/
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