SDNU 1537.Tiger Eat People

Description

As we all know, lls is the boss of SDNU ACM Training Team, so many team members may be afraid of him.
But this is not the case, lls is very kindly to us. To show his love to us, he plan to give all of us an "Accepted". The problem shows as follows:
There are n numbers .Each time you can choose a subset of them(may be empty), and then add them up.Count how many numbers can be represented in this way.

Input

The first line of the input contains an integer T, denoting the number of test cases.
In each test case, there is a single integers n in one line, as described above.
,

Output

For each test case, output one line contains a single integer, denoting the answer.

Sample Input

4
9
7
8
233

Sample Output

512
128
256
13803492693581127574869511724554050904902217944340773110325048447598592
#include <cstdio>
#include <iostream>
#include <queue>
#include <cstring>
#include <string>
#include <cmath>

using namespace std;

#define ll long long
#define maxn 1000000+8

int bit[maxn];

int main()
{
    int t;
    scanf("%d", &t);
    for(int i = 0; i < t; i++)
    {
        int n;
        scanf("%d", &n);
        int l = 0;
        bit[0] = 1;//相乘先把第一个数弄成0
        for(int i = 1; i <= n; i++)
        {
            int a = 0;
            for(int j = 0; j<= l;j++)
            {
                bit[j] = bit[j]*2+a;//计算2的次方
                a = 0;
                if(bit[j] >= 10)
                {
                    bit[j] -= 10;//将10位进到下一个
                    a++;
                }
            }
            if(a)//如果需要进位
                {
                    l++;
                    bit[l] = 1;//那这个大数据就向前进一位
                }
        }
        for(int i = l; i >= 0; i--)
        {
            printf("%d", bit[i]);
            bit[i] = 0;
        }
        printf("
");
    }
    return 0;
}
原文地址:https://www.cnblogs.com/RootVount/p/10328506.html