SDNU 1537.Tiger Eat People-优快云博客

本文详细解析了一道ACM竞赛中常见的子集求和问题，通过实例输入输出展示了如何计算给定数量的数字能以多少种方式相加得到的目标数。代码采用C++实现，利用位运算技巧高效解决了大规模数据处理问题。

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

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

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("\n");
    }
    return 0;
}