题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5703
Desert
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 177 Accepted Submission(s): 141
Problem Description
A tourist gets lost in the desert with n liters of water. He drinks positive integer units of water each day.
Write a program to calculate how many different ways the tourist can drink up the water.
Input
The first line contains the number of test cases T(T≤10).
Next T lines
contain the number n(1≤n≤1000000) for
each test case.
Output
Output consists of T lines.
Each line contains the binary number which represents number of different ways to finish up the water specified in the test case.
Sample Input
1 3
Sample Output
100 Hint 3 liters of water can be comsumed in four different ways show in the following. 1. 1 1 1 2. 1 2 3. 2 1 4. 3 If we write 4 in binary, it‘s 100.
Source
题目大意：一杯水有n的容量，问有多少种方法可以喝完。
提示给了3的水，就有四种方法喝完：1、1 1 1 2、1 2 3、2 1 4、3综上，一共是4种方法，4转换为二进制就是100
多次找规律之后，很明显就是2的n-1次方。
解题思路：方法数根据找到的规律很好计算，但是由于n太大，2的n次方特别大。
所以我们不能先算出方法数在进行二进制转换。
多次计算发现有这个规律。2的多少次方就有多少个0，所以先输出一个1，在接着输入0的个数就可以了。
详见代码。
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while (t--)
{
int n;
scanf("%d",&n);
printf ("1");
for (int i=0;i<n-1;i++)
printf ("0");
printf ("\n");
}
return 0;
}