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九度OJ 1095 2的幂次方
时间限制：1 秒
内存限制：32 兆
特殊判题：否
提交：772
解决：525
- 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)
#include<stdio.h>
#include<math.h>
void solve(int n)
{
int cnt=0;
while(pow(2,cnt)<=n)
cnt++;
cnt--;
if(cnt==0)
printf("2(0)");
else if(cnt==1)
printf("2");
else
{
printf("2(");
solve(cnt);
printf(")");
}
int num_r=n-pow(2,cnt);
if(num_r)
{
printf("+");
solve(num_r);
}
}
int main()
{
int n;
while(~scanf("%d",&n))
{
solve(n);
printf("\n");
}
return 0;
}
/**************************************************************
Problem: 1095
User: kirchhoff
Language: C
Result: Accepted
Time:0 ms
Memory:1004 kb
****************************************************************/九度OJ 1095 2的幂次方
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