Problem A CodeForces 560A
Description
A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimumunfortunate sum?
Input
The first line contains number n (1 ≤ n ≤ 1000) — the number of values of the banknotes that used in Geraldion.
The second line contains n distinct space-separated numbers a1, a2, ..., an (1 ≤ ai ≤ 106) — the values of the banknotes.
Output
Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print - 1.
Sample Input
5
1 2 3 4 5
-1
分析:
这题只用两种可能,就是输入有1与无1,有1时就找不到sum,所以输出-1;无1时有最小sum为1,即输出为1.
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int n,i;
while(scanf("%d",&n)==1&&n)
{
int f=0;
for(i=0;i<n;i++)
{
int x;
scanf("%d",&x);
if(x==1)
f=1;
}
if(!f)
printf("1\n");
else
printf("-1\n");
}
return 0;
}
Problem A CodeForces 560A的更多相关文章
- Problem - D - Codeforces Fix a Tree
Problem - D - Codeforces Fix a Tree 看完第一名的代码,顿然醒悟... 我可以把所有单独的点全部当成线,那么只有线和环. 如果全是线的话,直接线的条数-1,便是操作 ...
- Codeforces Round #439 (Div. 2) Problem E (Codeforces 869E) - 暴力 - 随机化 - 二维树状数组 - 差分
Adieu l'ami. Koyomi is helping Oshino, an acquaintance of his, to take care of an open space around ...
- Codeforces Round #439 (Div. 2) Problem C (Codeforces 869C) - 组合数学
— This is not playing but duty as allies of justice, Nii-chan! — Not allies but justice itself, Onii ...
- Codeforces Round #439 (Div. 2) Problem B (Codeforces 869B)
Even if the world is full of counterfeits, I still regard it as wonderful. Pile up herbs and incense ...
- Codeforces Round #439 (Div. 2) Problem A (Codeforces 869A) - 暴力
Rock... Paper! After Karen have found the deterministic winning (losing?) strategy for rock-paper-sc ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem F (Codeforces 831F) - 数论 - 暴力
题目传送门 传送门I 传送门II 传送门III 题目大意 求一个满足$d\sum_{i = 1}^{n} \left \lceil \frac{a_i}{d} \right \rceil - \sum ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem D (Codeforces 831D) - 贪心 - 二分答案 - 动态规划
There are n people and k keys on a straight line. Every person wants to get to the office which is l ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem E (Codeforces 831E) - 线段树 - 树状数组
Vasily has a deck of cards consisting of n cards. There is an integer on each of the cards, this int ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem C (Codeforces 831C) - 暴力 - 二分法
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain ...
随机推荐
- PSR-0的规范。
- 多路径(multi-path)安装测试实例
1.确保安装以下的包: device-mapper device-mapper-multipath [root@nticket1~]# rpm -qa "*device*" dev ...
- Windows 上远程访问 Unix 的 XWindow / XManager / X
准备 下载 putty - http://www.putty.org/ 安装 cygwin - http://cygwin.com/, 并添加 e.g. c:/cygwin/bin 到 Window ...
- hiho_1054_滑动解锁
题目大意 智能手机九点屏幕滑动解锁,如果给出某些连接线段,求出经过所有给出线段的合法的滑动解锁手势的总数.题目链接: 滑动解锁 题目分析 首先,尝试求解没有给定线段情况下,所有合法的路径的总数.可以使 ...
- iOS AVCaptureVideoDataOutputSampleBufferDelegate 录制视频
iOS AVCaptureVideoDataOutputSampleBufferDelegate 录制视频 应用场景: 使用AVFoundation提供的API, 我们可以从 AVCaptureVid ...
- C++模板特化
一 ."函数模板"与"模板函数" 下面几行代码就是一个"函数模板" template <class T> T abs(T x) ...
- linux笔记:linux系统安装-vmware虚拟机安装
vmware版本:vmware8(百度云里备份了安装程序VMware_Workstation_wmb.zip) vmware软件安装过程: 1.在百度云中下载安装程序压缩包VMware_Worksta ...
- IOC Container(服务容器)的工作机制
IOC Container 是laravel的一个核心内容,有了IOC Container在Laravel的强大表现,我们可以在Laravel中实现很大程度的代码维护性.(文档我是看的懵逼懵逼的(*^ ...
- Mybatis 学习-4
Category与Article双向一对多关联 (1)将CategoryDao进行实现 public class CategoryDaoImpl extends BaseDao<Category ...
- android入门之: SharedPreferences
读取数据: 保存数据: +++++++++++++++++++方法详解++++++++++++++++++++++++++++++ SharedPreferences综述: 使用getSharedPr ...