集训第四周（高效算法设计）H题 (贪心)

Description

 

Most financial institutions had become insolvent during financial crisis and went bankrupt or were bought by larger institutions, usually by banks. By the end of financial crisis of all the financial institutions only two banks still continue to operate. Financial markets had remained closed throughout the crisis and now regulators are gradually opening them. To prevent speculation and to gradually ramp up trading they will initially allow trading in only one financial instrument and the volume of trading will be limited to i<tex2html_verbatim_mark> contracts for i<tex2html_verbatim_mark> -th minute of market operation. Two banks had decided to cooperate with the government to kick-start the market operation. The boards of directors had agreed on trading volume for each minute of this first trading session. One bank will be buying a_i<tex2html_verbatim_mark> contracts ( 1a_ii<tex2html_verbatim_mark> ) during i<tex2html_verbatim_mark> -th minute ( 1in<tex2html_verbatim_mark> ), while the other one will be selling. They do not really care whether to buy or to sell, and the outside observer will only see the volume a_i<tex2html_verbatim_mark> of contracts traded per minute. However, they do not want to take any extra risk and want to have no position in the contract by the end of the trading session. Thus, if we define b_i = 1<tex2html_verbatim_mark> when the first bank is buying and b_i = - 1<tex2html_verbatim_mark> when the second one is buying (and the first one is selling), then the requirement for the trading session is that a_ib_i = 0<tex2html_verbatim_mark> . Your lucky team of three still works in the data center (due to the crisis, banks now share the data center and its personnel) and your task is to find such b_i<tex2html_verbatim_mark> or to report that this is impossible.

Input 

The input file contains several test cases, each of them as described below. The first line of the input contains the single integer number n<tex2html_verbatim_mark>( 1n100 000<tex2html_verbatim_mark> ). The second line of the input contains n<tex2html_verbatim_mark> integer numbers -- a_i<tex2html_verbatim_mark> ( 1a_ii<tex2html_verbatim_mark> ).

Output 

For each test case, the first line of the output must contain `` Yes'' if the trading session with specified volumes is possible and `` No'' otherwise. In the former option a second line must contain n<tex2html_verbatim_mark> numbers -- b_i<tex2html_verbatim_mark> .

Sample Input 

4
1 2 3 3
4
1 2 3 4

Sample Output 

No
Yes
1 -1 -1 1

给你一个数组，选择其中的一半的数变成负数，然后这个数组的和为0
首先来判断一下NO的情况，如果这个数组的和是一个奇数的话，那么你无论怎么选都不会有结果，然后如果你数组长度为1，那么就没得选了。
YES该怎么做？先从大到小排序，然后一路加下去（使和大于数组和一半的不能加），如果刚好和等于数组和的一半，那么你选择的那些数就是应该乘以-1的数了

#include"iostream"
#include"algorithm"
using namespace std;

const int maxn=100000+10;

int  f[maxn];
long long sum;
int n;

struct node
{
    int val,num;
}a[maxn];

bool cmp(struct node a1,struct node a2)
{
    return a1.val>a2.val;
}

bool Init()
{
    sum=0;
    for(int i=0;i<n;i++)
    {
        cin>>a[i].val;
        a[i].num=i;
        f[i]=-1;
        sum+=a[i].val;
    }
    if(sum%2) return false;
    else return true;
}

void Work()
{
    sort(a,a+n,cmp);
   long long cur=0;
    for(int i=0;i<n;i++)
    {
        if(cur+a[i].val<=sum/2)
        {
            cur+=a[i].val;
            f[a[i].num]=1;
            if(cur==sum/2) break;
        }
    }
}

void print()
{
    for(int i=0;i<n-1;i++)
    {
        if(i!=0) cout<<' ';
        cout<<f[i];
    }
    cout<<' '<<f[n-1]<<endl;
}

int main()
{
    while(cin>>n)
    {
        if(!Init()||n==1) cout<<"No"<<endl;
        else
        {
        cout<<"Yes"<<endl;
        Work();
        print();
        }
    }
    return 0;
}