题意:略.

析:不解释,水题。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
using namespace std :: tr1; typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 30 + 5;
const int mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
} int main(){
int T; cin >> T;
while(T--){
scanf("%d", &n);
printf("%d\n", (1<<n)-1);
}
return 0;
}

UVaLive 6591 && Gym 100299L Bus (水题)的更多相关文章

  1. Codeforces Gym 100531G Grave 水题

    Problem G. Grave 题目连接: http://codeforces.com/gym/100531/attachments Description Gerard develops a Ha ...

  2. Codeforces Gym 100286I iSharp 水题

    Problem I. iSharpTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/ ...

  3. CodeForces Gym 100685C Cinderella (水题)

    题意:给定 n 个杯子,里面有不同体积的水,然后问你要把所有的杯子的水的体积都一样,至少要倒少多少个杯子. 析:既然最后都一样,那么先求平均数然后再数一下,哪个杯子的开始的体积就大于平均数,这是一定要 ...

  4. Proving Equivalences UVALive - 4287(强连通分量 水题)

    就是统计入度为0 的点 和 出度为0 的点  输出 大的那一个,, 若图中只有一个强连通分量 则输出0即可 和https://www.cnblogs.com/WTSRUVF/p/9301096.htm ...

  5. Educational Codeforces Round 11 B. Seating On Bus 水题

    B. Seating On Bus 题目连接: http://www.codeforces.com/contest/660/problem/B Description Consider 2n rows ...

  6. Two Operations Gym - 102263M 优先队列水题

    Two Operations Gym - 102263M Ayoub has a string SS consists of only lower case Latin letters, and he ...

  7. UVALive 4192 Close Enough Computations 水题

    Close Enough Computations 题目连接: https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge& ...

  8. UVALive 7275 Dice Cup (水题)

    Dice Cup 题目链接: http://acm.hust.edu.cn/vjudge/contest/127406#problem/D Description In many table-top ...

  9. Gym 100531G Grave(水题)

    题意:给定一个大矩形,再给定在一个小矩形,然后给定一个新矩形的长和高,问你能不能把这个新矩形放到大矩形里,并且不与小矩形相交. 析:直接判定小矩形的上下左右四个方向,能不能即可. 代码如下: #pra ...

随机推荐

  1. 【Todo】UDP P2P打洞原理

    参考以下两篇文章: https://my.oschina.net/ososchina/blog/369206 http://m.blog.csdn.net/article/details?id=666 ...

  2. 安装ftp服务器

    Linux安装ftp组件 1  安装vsftpd组件 安装完后,有/etc/vsftpd/vsftpd.conf文件,是vsftp的配置文件. [root@bogon ~]# yum -y insta ...

  3. leetCode 65.Valid Number (有效数字)

    Valid Number  Validate if a given string is numeric. Some examples: "0" => true " ...

  4. java8 stream sorted

    1.对象类型配列 List<Person> list = Arrays.asList( new Person(22, "shaomch", "man" ...

  5. ime-mode:disabled (用css实现关闭文本框输入法)

    css 之 ime-mode语法:ime-mode : auto | active | inactive | disabled取值:auto : 默认值.不影响ime的状态.与不指定 ime-mode ...

  6. hud 1465、2049、2045 (递推)[含简单C(n,m) n!的写法]

    C - 不容易系列之一 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit  ...

  7. HDU 1423 Greatest Common Increasing Subsequence(LICS入门,只要求出最长数)

    Greatest Common Increasing Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536 ...

  8. 全卷积神经网络FCN理解

    论文地址:https://people.eecs.berkeley.edu/~jonlong/long_shelhamer_fcn.pdf 这篇论文使用全卷积神经网络来做语义上的图像分割,开创了这一领 ...

  9. label 对齐

    <label for="username">用户名</label> <input type="text" id="use ...

  10. Java Unit Testing - JUnit & TestNG

    转自https://www3.ntu.edu.sg/home/ehchua/programming/java/JavaUnitTesting.html yet another insignifican ...