Codeforces Round #238 (Div. 2) D. Toy Sum 暴搜
题目链接:
题目
D. Toy Sum
time limit per test:1 second
memory limit per test:256 megabytes
问题描述
Little Chris is very keen on his toy blocks. His teacher, however, wants Chris to solve more problems, so he decided to play a trick on Chris.
There are exactly s blocks in Chris's set, each block has a unique number from 1 to s. Chris's teacher picks a subset of blocks X and keeps it to himself. He will give them back only if Chris can pick such a non-empty subset Y from the remaining blocks, that the equality holds:
"Are you kidding me?", asks Chris.
For example, consider a case where s = 8 and Chris's teacher took the blocks with numbers 1, 4 and 5. One way for Chris to choose a set is to pick the blocks with numbers 3 and 6, see figure. Then the required sums would be equal: (1 - 1) + (4 - 1) + (5 - 1) = (8 - 3) + (8 - 6) = 7.
However, now Chris has exactly s = 106 blocks. Given the set X of blocks his teacher chooses, help Chris to find the required set Y!
输入
The first line of input contains a single integer n (1 ≤ n ≤ 5·105), the number of blocks in the set X. The next line contains n distinct space-separated integers x1, x2, ..., xn (1 ≤ xi ≤ 106), the numbers of the blocks in X.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
输出
In the first line of output print a single integer m (1 ≤ m ≤ 106 - n), the number of blocks in the set Y. In the next line output m distinct space-separated integers y1, y2, ..., ym (1 ≤ yi ≤ 106), such that the required equality holds. The sets X and Y should not intersect, i.e. xi ≠ yj for all i, j (1 ≤ i ≤ n; 1 ≤ j ≤ m). It is guaranteed that at least one solution always exists. If there are multiple solutions, output any of them.
样例
input
3
1 4 5
output
2
999993 1000000
input
1
1
output
1
1000000
题意
给你一个大小为n的X集合x1,x2,...,xn,问你能不能选定一个集合大小S(S<=10^6),使得从S包含X,并且从属于S但不属于X的元素组成的集合里,选出一个大小为k的非空子集Y,使得等式sigma(xi-1)==sigma(s-yj),其中1<=i<=n,1<=j<=k。
题解
1、暴搜,在属于S但不属于X的元素组成的集合里暴搜出一个满足答案的解。跑了717ms。(暴搜方案,每个数选或不选)
#include<iostream>
#include<cstdio>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;
typedef __int64 LL;
const int maxn=5e5+10;
const int INF=1e6;
int arr[maxn];
int n;
map<int,int> mp;
vector<int> ans;
vector<int> re;
bool dfs(int cur,LL sum){
if(sum==0) return true;
int pos=lower_bound(re.begin(),re.end(),sum)-re.begin();
if(pos>cur) pos=cur;
if(pos<=0) return false;
ans.push_back(INF-re[pos]);
if(dfs(pos-1,sum-re[pos])) return true;
ans.pop_back();
if(dfs(pos-1,sum)) return true;
return false;
}
int main(){
scanf("%d",&n);
LL sum=0;
for(int i=0;i<n;i++){
int x; scanf("%d",&x); mp[x]=1;
sum+=x-1;
}
re.clear();
re.push_back(-1);
for(int i=1;i<=INF;i++){
if(!mp[i]){
re.push_back(INF-i);
}
}
sort(re.begin(),re.end());
dfs(re.size()-1,sum);
if(!mp[INF]) ans.push_back(INF);
printf("%d\n",ans.size());
for(int i=0;i<ans.size()-1;i++) printf("%d ",ans[i]);
printf("%d\n",ans[ans.size()-1]);
return 0;
}
2、找对称的数,对于x,有y=S+1-i,使得x-1=S-y;所以对于xi,如果与它对称的数yi不在X里面,则吧yi放到答案里面,如果在,则这两个数相加为S,我们只要另找一个数不在X里面,并且它对称的数也不在X里面,然后吧这两个数都加到答案里面就可以了。
#include<iostream>
#include<vector>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 1e6 + 10;
const int S = 1e6;
int used[maxn];
int main() {
memset(used, 0, sizeof(used));
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
int x; scanf("%d", &x); used[x] = 1;
}
vector<int> ans;
//cnt统计一个数和与它对称的数都在X里面的数对
int cnt = 0;
for (int i = 1; i <= S; i++) {
if (used[i]) {
int j = S + 1 - i;
if (!used[j]) {
ans.push_back(j);
used[j] = 1;
}
else {
if(i<=j) cnt++;
}
}
}
for (int i = 1; i <= S&&cnt>0; i++) {
if (!used[i]) {
int j = S + 1 - i;
ans.push_back(i);
ans.push_back(j);
used[i] = used[j] = 1;
cnt--;
}
}
sort(ans.begin(), ans.end());
printf("%d\n", ans.size());
for (int i = 0; i < ans.size() - 1; i++) printf("%d ", ans[i]);
printf("%d\n",ans[ans.size()-1]);
return 0;
}
Codeforces Round #238 (Div. 2) D. Toy Sum 暴搜的更多相关文章
- Codeforces Round #238 (Div. 2) D. Toy Sum
D. Toy Sum time limit per test:1 second memory limit per test:256 megabytes input:standard input o ...
- Codeforces Round #238 (Div. 2) D. Toy Sum(想法题)
传送门 Description Little Chris is very keen on his toy blocks. His teacher, however, wants Chris to s ...
- 水题 Codeforces Round #303 (Div. 2) A. Toy Cars
题目传送门 /* 题意:5种情况对应对应第i或j辆车翻了没 水题:其实就看对角线的上半边就可以了,vis判断,可惜WA了一次 3: if both cars turned over during th ...
- Codeforces Round #556 (Div. 2) - C. Prefix Sum Primes(思维)
Problem Codeforces Round #556 (Div. 2) - D. Three Religions Time Limit: 1000 mSec Problem Descripti ...
- Codeforces Round #238 (Div. 1)
感觉这场题目有种似曾相识感觉,C题还没看,日后补上.一定要坚持做下去. A Unusual Product 题意: 给定一个n*n的01矩阵,3种操作, 1 i 将第i行翻转 2 i 将第i列翻转 3 ...
- Codeforces Codeforces Round #319 (Div. 2) B. Modulo Sum 背包dp
B. Modulo Sum Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/577/problem/ ...
- Codeforces Round #303 (Div. 2) A. Toy Cars 水题
A. Toy Cars Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/545/problem ...
- Codeforces Round #344 (Div. 2) E. Product Sum 维护凸壳
E. Product Sum 题目连接: http://www.codeforces.com/contest/631/problem/E Description Blake is the boss o ...
- Codeforces Round #232 (Div. 2) D. On Sum of Fractions
D. On Sum of Fractions Let's assume that v(n) is the largest prime number, that does not exceed n; u ...
随机推荐
- asp连接SQL数据库的代码
connstr="driver={SQL Server};server=(local);uid=sa;pwd=sa;database=Your database" 语法介绍: 1. ...
- UIView的frame的扩展分类,轻松取出x、y、height、width等值
一.引言: 在ios开发中,就界面搭建.控件布局时,都会很恶心的通过很长的代码才能取出控件的x.y.height.width等值,大大降低了开发效率.那为了省略这些恶心的步骤,小编在这里给UIView ...
- UI5_HomeWork
// // View.h // UI5_HomeWork // // Created by zhangxueming on 15/7/2. // Copyright (c) 2015年 zhangxu ...
- 异常getaddrinfo enotfound
在看NodeJS开发指南这本书时,书中的一个例子,讲解http.request的.代码如下: var http = require('http'); var querystring = require ...
- 字符集转换: Ansi - Unicode
字符集转换: Ansi - Unicode wstring AnsiToUnicode (const string& strSrc ) { /*!< 分配目标空间 */ ,strSrc. ...
- 30类css选择器
大概大家都知道id,class以及descendant选择器,并且整体都在使用它们,那么你正在错误拥有更大级别的灵活性的选择方式.这篇文章里面提到的大部分选择器都是在CSS3标准下的,所以它们只能在相 ...
- 杭电ACM2098--分拆素数和
题目:http://acm.hdu.edu.cn/showproblem.php?pid=2098 这是源码.其实我本不想拿出源码,毕竟源码很容易被复制. 我这里刚开始出错的地方有 0_0_12811 ...
- C#调用C、C++结构体数组的方法总结
一个客户要使用C#调用我们用C++开发的一个动态链接库,本来我没有C#的开发经验,就随便写了一个例程.以为很简单就可以搞定,没想到客户开发的过程中遇到了不少问题,最困难的就是用C#调用C++接口中的自 ...
- 必须知道的.NET FrameWork
这部分内容呢,对普通开发者来说,意义不是特别大.但是作为一个有情怀的开发者,应该追根溯源,懂一些自己用的工具, 的粗浅知识.并且这部分内容可以用来个HR吹嘘一下自己的情怀.基本就是这样. .NET生成 ...
- javascript框架库API入口
underscorejs : http://learning.github.io/underscore/