PAT-1144(The Missing Number)set的使用,简单题
The Missing Number
PAT-1144
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cstdio>
#include<sstream>
#include<set>
using namespace std;
const int maxn=100005;
set<int>se;
int main(){
int n;
cin>>n;
for(int i=0;i<n;i++){
int a;
cin>>a;
if(a>0)
se.insert(a);
}
if(se.empty()){
cout<<1<<endl;
return 0;
}
set<int>::iterator it=se.begin();
int pre=*it;
it++;
for(;it!=se.end();it++){
int now=*it;
// cout<<now<<endl;
if(now!=pre+1){
cout<<pre+1<<endl;
return 0;
}else pre=now;
}
cout<<pre+1<<endl;
return 0;
}
PAT-1144(The Missing Number)set的使用,简单题的更多相关文章
- PAT 1144 The Missing Number
1144 The Missing Number (20 分) Given N integers, you are supposed to find the smallest positive in ...
- PAT 1144 The Missing Number[简单]
1144 The Missing Number(20 分) Given N integers, you are supposed to find the smallest positive integ ...
- [PAT] 1144 The Missing Number(20 分)
1144 The Missing Number(20 分) Given N integers, you are supposed to find the smallest positive integ ...
- PAT 甲级 1144 The Missing Number (20 分)(简单,最后一个测试点没过由于开的数组没必要大于N)
1144 The Missing Number (20 分) Given N integers, you are supposed to find the smallest positive in ...
- PAT(A) 1144 The Missing Number(C)统计
题目链接:1144 The Missing Number (20 point(s)) Description Given N integers, you are supposed to find th ...
- PAT 甲级 1144 The Missing Number
https://pintia.cn/problem-sets/994805342720868352/problems/994805343463260160 Given N integers, you ...
- PAT A1144 The Missing Number (20 分)——set
Given N integers, you are supposed to find the smallest positive integer that is NOT in the given li ...
- 1144 The Missing Number (20 分)
Given N integers, you are supposed to find the smallest positive integer that is NOT in the given li ...
- 1144. The Missing Number (20)
Given N integers, you are supposed to find the smallest positive integer that is NOT in the given li ...
- 1144 The Missing Number
Given N integers, you are supposed to find the smallest positive integer that is NOT in the given li ...
随机推荐
- Codeforces Round #602 Div2 D1. Optimal Subsequences (Easy Version)
题意:给你一个数组a,询问m次,每次返回长度为k的和最大的子序列(要求字典序最小)的pos位置上的数字. 题解:和最大的子序列很简单,排个序就行,但是题目要求字典序最小,那我们在刚开始的时候先记录每个 ...
- 微服务架构Day02-SpringBoot日志slf4j
日志框架 日志门面(接口,日志抽象层 ) 日志实现 JCL(Jakarta Commons Logging).slf4j(Simple Logging Facade for Java).jboss-l ...
- vue 二级子路由跳转不了 bug
vue 二级子路由跳转不了 bug @click.prevent 阻止原生事件的冒泡 <li class="tools-hover-box-list-item" v-for= ...
- vue & npm & components & plugins
vue & npm & components & plugins how to publish an vue ui component to npm? https://www. ...
- css & background-image & full page width & background-size
css & background-image & full page width & background-size https://css-tricks.com/perfec ...
- node.js 如何处理一个很大的文件
node.js 如何处理一个很大的文件 思路 arraybuffer 数据分段 时间分片 多线程 web workers sevice workers node.js 如何处理一个很大的文件 http ...
- SameSite cookies explained
SameSite cookies explained
- react hooks & component will unmount & useEffect & clear up
react hooks & component will unmount & useEffect & clear up useEffect & return === u ...
- 2021-2-19:请问你知道 Java 如何高性能操作文件么?
一般高性能的涉及到存储框架,例如 RocketMQ,Kafka 这种消息队列,存储日志的时候,都是通过 Java File MMAP 实现的,那么什么是 Java File MMAP 呢? 什么是 J ...
- www.yimitv.cc免费观看2020最新电影、电视剧、综艺栏目
神奇的微信公众号 '德佑小站', 可以看最新上映电影.看小说.看直播!重要的是免费,csdn已加速. 壹米影视:www.yimitv.cc 德佑小说:www.deyouxs.cc